/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


--------------------------------------------------------------------------------


YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Termination w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) Overlay + Local Confluence [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QReductionProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QReductionProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (20) YES
    (21) QDP
        (22) UsableRulesProof [EQUIVALENT, 0 ms]
        (23) QDP
        (24) QReductionProof [EQUIVALENT, 0 ms]
        (25) QDP
        (26) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (27) YES
    (28) QDP
        (29) UsableRulesProof [EQUIVALENT, 0 ms]
        (30) QDP
        (31) QReductionProof [EQUIVALENT, 0 ms]
        (32) QDP
        (33) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (34) YES
    (35) QDP
        (36) UsableRulesProof [EQUIVALENT, 0 ms]
        (37) QDP
        (38) QReductionProof [EQUIVALENT, 0 ms]
        (39) QDP
        (40) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (41) YES
    (42) QDP
        (43) UsableRulesProof [EQUIVALENT, 0 ms]
        (44) QDP
        (45) QReductionProof [EQUIVALENT, 0 ms]
        (46) QDP
        (47) Induction-Processor [SOUND, 47 ms]
        (48) AND
            (49) QDP
                (50) UsableRulesProof [EQUIVALENT, 0 ms]
                (51) QDP
                (52) QReductionProof [EQUIVALENT, 0 ms]
                (53) QDP
                (54) Induction-Processor [SOUND, 26 ms]
                (55) AND
                    (56) QDP
                        (57) PisEmptyProof [EQUIVALENT, 0 ms]
                        (58) YES
                    (59) QTRS
                        (60) QTRSRRRProof [EQUIVALENT, 69 ms]
                        (61) QTRS
                        (62) QTRSRRRProof [EQUIVALENT, 0 ms]
                        (63) QTRS
                        (64) RisEmptyProof [EQUIVALENT, 0 ms]
                        (65) YES
            (66) QTRS
                (67) QTRSRRRProof [EQUIVALENT, 105 ms]
                (68) QTRS
                (69) QTRSRRRProof [EQUIVALENT, 0 ms]
                (70) QTRS
                (71) QTRSRRRProof [EQUIVALENT, 0 ms]
                (72) QTRS
                (73) RisEmptyProof [EQUIVALENT, 0 ms]
                (74) YES


----------------------------------------

(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   qsort(nil) -> nil
   qsort(cons(x, xs)) -> append(qsort(filterlow(last(cons(x, xs)), cons(x, xs))), cons(last(cons(x, xs)), qsort(filterhigh(last(cons(x, xs)), cons(x, xs)))))
   filterlow(n, nil) -> nil
   filterlow(n, cons(x, xs)) -> if1(ge(n, x), n, x, xs)
   if1(true, n, x, xs) -> filterlow(n, xs)
   if1(false, n, x, xs) -> cons(x, filterlow(n, xs))
   filterhigh(n, nil) -> nil
   filterhigh(n, cons(x, xs)) -> if2(ge(x, n), n, x, xs)
   if2(true, n, x, xs) -> filterhigh(n, xs)
   if2(false, n, x, xs) -> cons(x, filterhigh(n, xs))
   ge(x, 0) -> true
   ge(0, s(x)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   append(nil, ys) -> ys
   append(cons(x, xs), ys) -> cons(x, append(xs, ys))
   last(nil) -> 0
   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))

Q is empty.

----------------------------------------

(1) Overlay + Local Confluence (EQUIVALENT)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
----------------------------------------

(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   qsort(nil) -> nil
   qsort(cons(x, xs)) -> append(qsort(filterlow(last(cons(x, xs)), cons(x, xs))), cons(last(cons(x, xs)), qsort(filterhigh(last(cons(x, xs)), cons(x, xs)))))
   filterlow(n, nil) -> nil
   filterlow(n, cons(x, xs)) -> if1(ge(n, x), n, x, xs)
   if1(true, n, x, xs) -> filterlow(n, xs)
   if1(false, n, x, xs) -> cons(x, filterlow(n, xs))
   filterhigh(n, nil) -> nil
   filterhigh(n, cons(x, xs)) -> if2(ge(x, n), n, x, xs)
   if2(true, n, x, xs) -> filterhigh(n, xs)
   if2(false, n, x, xs) -> cons(x, filterhigh(n, xs))
   ge(x, 0) -> true
   ge(0, s(x)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   append(nil, ys) -> ys
   append(cons(x, xs), ys) -> cons(x, append(xs, ys))
   last(nil) -> 0
   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))

The set Q consists of the following terms:

   qsort(nil)
   qsort(cons(x0, x1))
   filterlow(x0, nil)
   filterlow(x0, cons(x1, x2))
   if1(true, x0, x1, x2)
   if1(false, x0, x1, x2)
   filterhigh(x0, nil)
   filterhigh(x0, cons(x1, x2))
   if2(true, x0, x1, x2)
   if2(false, x0, x1, x2)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   append(nil, ys)
   append(cons(x0, x1), ys)
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))


----------------------------------------

(3) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   QSORT(cons(x, xs)) -> APPEND(qsort(filterlow(last(cons(x, xs)), cons(x, xs))), cons(last(cons(x, xs)), qsort(filterhigh(last(cons(x, xs)), cons(x, xs)))))
   QSORT(cons(x, xs)) -> QSORT(filterlow(last(cons(x, xs)), cons(x, xs)))
   QSORT(cons(x, xs)) -> FILTERLOW(last(cons(x, xs)), cons(x, xs))
   QSORT(cons(x, xs)) -> LAST(cons(x, xs))
   QSORT(cons(x, xs)) -> QSORT(filterhigh(last(cons(x, xs)), cons(x, xs)))
   QSORT(cons(x, xs)) -> FILTERHIGH(last(cons(x, xs)), cons(x, xs))
   FILTERLOW(n, cons(x, xs)) -> IF1(ge(n, x), n, x, xs)
   FILTERLOW(n, cons(x, xs)) -> GE(n, x)
   IF1(true, n, x, xs) -> FILTERLOW(n, xs)
   IF1(false, n, x, xs) -> FILTERLOW(n, xs)
   FILTERHIGH(n, cons(x, xs)) -> IF2(ge(x, n), n, x, xs)
   FILTERHIGH(n, cons(x, xs)) -> GE(x, n)
   IF2(true, n, x, xs) -> FILTERHIGH(n, xs)
   IF2(false, n, x, xs) -> FILTERHIGH(n, xs)
   GE(s(x), s(y)) -> GE(x, y)
   APPEND(cons(x, xs), ys) -> APPEND(xs, ys)
   LAST(cons(x, cons(y, xs))) -> LAST(cons(y, xs))

The TRS R consists of the following rules:

   qsort(nil) -> nil
   qsort(cons(x, xs)) -> append(qsort(filterlow(last(cons(x, xs)), cons(x, xs))), cons(last(cons(x, xs)), qsort(filterhigh(last(cons(x, xs)), cons(x, xs)))))
   filterlow(n, nil) -> nil
   filterlow(n, cons(x, xs)) -> if1(ge(n, x), n, x, xs)
   if1(true, n, x, xs) -> filterlow(n, xs)
   if1(false, n, x, xs) -> cons(x, filterlow(n, xs))
   filterhigh(n, nil) -> nil
   filterhigh(n, cons(x, xs)) -> if2(ge(x, n), n, x, xs)
   if2(true, n, x, xs) -> filterhigh(n, xs)
   if2(false, n, x, xs) -> cons(x, filterhigh(n, xs))
   ge(x, 0) -> true
   ge(0, s(x)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   append(nil, ys) -> ys
   append(cons(x, xs), ys) -> cons(x, append(xs, ys))
   last(nil) -> 0
   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))

The set Q consists of the following terms:

   qsort(nil)
   qsort(cons(x0, x1))
   filterlow(x0, nil)
   filterlow(x0, cons(x1, x2))
   if1(true, x0, x1, x2)
   if1(false, x0, x1, x2)
   filterhigh(x0, nil)
   filterhigh(x0, cons(x1, x2))
   if2(true, x0, x1, x2)
   if2(false, x0, x1, x2)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   append(nil, ys)
   append(cons(x0, x1), ys)
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 6 SCCs with 6 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

----------------------------------------

(7)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   LAST(cons(x, cons(y, xs))) -> LAST(cons(y, xs))

The TRS R consists of the following rules:

   qsort(nil) -> nil
   qsort(cons(x, xs)) -> append(qsort(filterlow(last(cons(x, xs)), cons(x, xs))), cons(last(cons(x, xs)), qsort(filterhigh(last(cons(x, xs)), cons(x, xs)))))
   filterlow(n, nil) -> nil
   filterlow(n, cons(x, xs)) -> if1(ge(n, x), n, x, xs)
   if1(true, n, x, xs) -> filterlow(n, xs)
   if1(false, n, x, xs) -> cons(x, filterlow(n, xs))
   filterhigh(n, nil) -> nil
   filterhigh(n, cons(x, xs)) -> if2(ge(x, n), n, x, xs)
   if2(true, n, x, xs) -> filterhigh(n, xs)
   if2(false, n, x, xs) -> cons(x, filterhigh(n, xs))
   ge(x, 0) -> true
   ge(0, s(x)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   append(nil, ys) -> ys
   append(cons(x, xs), ys) -> cons(x, append(xs, ys))
   last(nil) -> 0
   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))

The set Q consists of the following terms:

   qsort(nil)
   qsort(cons(x0, x1))
   filterlow(x0, nil)
   filterlow(x0, cons(x1, x2))
   if1(true, x0, x1, x2)
   if1(false, x0, x1, x2)
   filterhigh(x0, nil)
   filterhigh(x0, cons(x1, x2))
   if2(true, x0, x1, x2)
   if2(false, x0, x1, x2)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   append(nil, ys)
   append(cons(x0, x1), ys)
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(8) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(9)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   LAST(cons(x, cons(y, xs))) -> LAST(cons(y, xs))

R is empty.
The set Q consists of the following terms:

   qsort(nil)
   qsort(cons(x0, x1))
   filterlow(x0, nil)
   filterlow(x0, cons(x1, x2))
   if1(true, x0, x1, x2)
   if1(false, x0, x1, x2)
   filterhigh(x0, nil)
   filterhigh(x0, cons(x1, x2))
   if2(true, x0, x1, x2)
   if2(false, x0, x1, x2)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   append(nil, ys)
   append(cons(x0, x1), ys)
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(10) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   qsort(nil)
   qsort(cons(x0, x1))
   filterlow(x0, nil)
   filterlow(x0, cons(x1, x2))
   if1(true, x0, x1, x2)
   if1(false, x0, x1, x2)
   filterhigh(x0, nil)
   filterhigh(x0, cons(x1, x2))
   if2(true, x0, x1, x2)
   if2(false, x0, x1, x2)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   append(nil, ys)
   append(cons(x0, x1), ys)
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))


----------------------------------------

(11)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   LAST(cons(x, cons(y, xs))) -> LAST(cons(y, xs))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(12) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*LAST(cons(x, cons(y, xs))) -> LAST(cons(y, xs))
The graph contains the following edges 1 > 1


----------------------------------------

(13)
YES

----------------------------------------

(14)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   APPEND(cons(x, xs), ys) -> APPEND(xs, ys)

The TRS R consists of the following rules:

   qsort(nil) -> nil
   qsort(cons(x, xs)) -> append(qsort(filterlow(last(cons(x, xs)), cons(x, xs))), cons(last(cons(x, xs)), qsort(filterhigh(last(cons(x, xs)), cons(x, xs)))))
   filterlow(n, nil) -> nil
   filterlow(n, cons(x, xs)) -> if1(ge(n, x), n, x, xs)
   if1(true, n, x, xs) -> filterlow(n, xs)
   if1(false, n, x, xs) -> cons(x, filterlow(n, xs))
   filterhigh(n, nil) -> nil
   filterhigh(n, cons(x, xs)) -> if2(ge(x, n), n, x, xs)
   if2(true, n, x, xs) -> filterhigh(n, xs)
   if2(false, n, x, xs) -> cons(x, filterhigh(n, xs))
   ge(x, 0) -> true
   ge(0, s(x)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   append(nil, ys) -> ys
   append(cons(x, xs), ys) -> cons(x, append(xs, ys))
   last(nil) -> 0
   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))

The set Q consists of the following terms:

   qsort(nil)
   qsort(cons(x0, x1))
   filterlow(x0, nil)
   filterlow(x0, cons(x1, x2))
   if1(true, x0, x1, x2)
   if1(false, x0, x1, x2)
   filterhigh(x0, nil)
   filterhigh(x0, cons(x1, x2))
   if2(true, x0, x1, x2)
   if2(false, x0, x1, x2)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   append(nil, ys)
   append(cons(x0, x1), ys)
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(15) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(16)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   APPEND(cons(x, xs), ys) -> APPEND(xs, ys)

R is empty.
The set Q consists of the following terms:

   qsort(nil)
   qsort(cons(x0, x1))
   filterlow(x0, nil)
   filterlow(x0, cons(x1, x2))
   if1(true, x0, x1, x2)
   if1(false, x0, x1, x2)
   filterhigh(x0, nil)
   filterhigh(x0, cons(x1, x2))
   if2(true, x0, x1, x2)
   if2(false, x0, x1, x2)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   append(nil, ys)
   append(cons(x0, x1), ys)
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(17) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   qsort(nil)
   qsort(cons(x0, x1))
   filterlow(x0, nil)
   filterlow(x0, cons(x1, x2))
   if1(true, x0, x1, x2)
   if1(false, x0, x1, x2)
   filterhigh(x0, nil)
   filterhigh(x0, cons(x1, x2))
   if2(true, x0, x1, x2)
   if2(false, x0, x1, x2)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   append(nil, ys)
   append(cons(x0, x1), ys)
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))


----------------------------------------

(18)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   APPEND(cons(x, xs), ys) -> APPEND(xs, ys)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(19) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*APPEND(cons(x, xs), ys) -> APPEND(xs, ys)
The graph contains the following edges 1 > 1, 2 >= 2


----------------------------------------

(20)
YES

----------------------------------------

(21)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   GE(s(x), s(y)) -> GE(x, y)

The TRS R consists of the following rules:

   qsort(nil) -> nil
   qsort(cons(x, xs)) -> append(qsort(filterlow(last(cons(x, xs)), cons(x, xs))), cons(last(cons(x, xs)), qsort(filterhigh(last(cons(x, xs)), cons(x, xs)))))
   filterlow(n, nil) -> nil
   filterlow(n, cons(x, xs)) -> if1(ge(n, x), n, x, xs)
   if1(true, n, x, xs) -> filterlow(n, xs)
   if1(false, n, x, xs) -> cons(x, filterlow(n, xs))
   filterhigh(n, nil) -> nil
   filterhigh(n, cons(x, xs)) -> if2(ge(x, n), n, x, xs)
   if2(true, n, x, xs) -> filterhigh(n, xs)
   if2(false, n, x, xs) -> cons(x, filterhigh(n, xs))
   ge(x, 0) -> true
   ge(0, s(x)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   append(nil, ys) -> ys
   append(cons(x, xs), ys) -> cons(x, append(xs, ys))
   last(nil) -> 0
   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))

The set Q consists of the following terms:

   qsort(nil)
   qsort(cons(x0, x1))
   filterlow(x0, nil)
   filterlow(x0, cons(x1, x2))
   if1(true, x0, x1, x2)
   if1(false, x0, x1, x2)
   filterhigh(x0, nil)
   filterhigh(x0, cons(x1, x2))
   if2(true, x0, x1, x2)
   if2(false, x0, x1, x2)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   append(nil, ys)
   append(cons(x0, x1), ys)
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(22) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(23)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   GE(s(x), s(y)) -> GE(x, y)

R is empty.
The set Q consists of the following terms:

   qsort(nil)
   qsort(cons(x0, x1))
   filterlow(x0, nil)
   filterlow(x0, cons(x1, x2))
   if1(true, x0, x1, x2)
   if1(false, x0, x1, x2)
   filterhigh(x0, nil)
   filterhigh(x0, cons(x1, x2))
   if2(true, x0, x1, x2)
   if2(false, x0, x1, x2)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   append(nil, ys)
   append(cons(x0, x1), ys)
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(24) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   qsort(nil)
   qsort(cons(x0, x1))
   filterlow(x0, nil)
   filterlow(x0, cons(x1, x2))
   if1(true, x0, x1, x2)
   if1(false, x0, x1, x2)
   filterhigh(x0, nil)
   filterhigh(x0, cons(x1, x2))
   if2(true, x0, x1, x2)
   if2(false, x0, x1, x2)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   append(nil, ys)
   append(cons(x0, x1), ys)
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))


----------------------------------------

(25)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   GE(s(x), s(y)) -> GE(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(26) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*GE(s(x), s(y)) -> GE(x, y)
The graph contains the following edges 1 > 1, 2 > 2


----------------------------------------

(27)
YES

----------------------------------------

(28)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF2(true, n, x, xs) -> FILTERHIGH(n, xs)
   FILTERHIGH(n, cons(x, xs)) -> IF2(ge(x, n), n, x, xs)
   IF2(false, n, x, xs) -> FILTERHIGH(n, xs)

The TRS R consists of the following rules:

   qsort(nil) -> nil
   qsort(cons(x, xs)) -> append(qsort(filterlow(last(cons(x, xs)), cons(x, xs))), cons(last(cons(x, xs)), qsort(filterhigh(last(cons(x, xs)), cons(x, xs)))))
   filterlow(n, nil) -> nil
   filterlow(n, cons(x, xs)) -> if1(ge(n, x), n, x, xs)
   if1(true, n, x, xs) -> filterlow(n, xs)
   if1(false, n, x, xs) -> cons(x, filterlow(n, xs))
   filterhigh(n, nil) -> nil
   filterhigh(n, cons(x, xs)) -> if2(ge(x, n), n, x, xs)
   if2(true, n, x, xs) -> filterhigh(n, xs)
   if2(false, n, x, xs) -> cons(x, filterhigh(n, xs))
   ge(x, 0) -> true
   ge(0, s(x)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   append(nil, ys) -> ys
   append(cons(x, xs), ys) -> cons(x, append(xs, ys))
   last(nil) -> 0
   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))

The set Q consists of the following terms:

   qsort(nil)
   qsort(cons(x0, x1))
   filterlow(x0, nil)
   filterlow(x0, cons(x1, x2))
   if1(true, x0, x1, x2)
   if1(false, x0, x1, x2)
   filterhigh(x0, nil)
   filterhigh(x0, cons(x1, x2))
   if2(true, x0, x1, x2)
   if2(false, x0, x1, x2)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   append(nil, ys)
   append(cons(x0, x1), ys)
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(29) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(30)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF2(true, n, x, xs) -> FILTERHIGH(n, xs)
   FILTERHIGH(n, cons(x, xs)) -> IF2(ge(x, n), n, x, xs)
   IF2(false, n, x, xs) -> FILTERHIGH(n, xs)

The TRS R consists of the following rules:

   ge(x, 0) -> true
   ge(0, s(x)) -> false
   ge(s(x), s(y)) -> ge(x, y)

The set Q consists of the following terms:

   qsort(nil)
   qsort(cons(x0, x1))
   filterlow(x0, nil)
   filterlow(x0, cons(x1, x2))
   if1(true, x0, x1, x2)
   if1(false, x0, x1, x2)
   filterhigh(x0, nil)
   filterhigh(x0, cons(x1, x2))
   if2(true, x0, x1, x2)
   if2(false, x0, x1, x2)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   append(nil, ys)
   append(cons(x0, x1), ys)
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(31) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   qsort(nil)
   qsort(cons(x0, x1))
   filterlow(x0, nil)
   filterlow(x0, cons(x1, x2))
   if1(true, x0, x1, x2)
   if1(false, x0, x1, x2)
   filterhigh(x0, nil)
   filterhigh(x0, cons(x1, x2))
   if2(true, x0, x1, x2)
   if2(false, x0, x1, x2)
   append(nil, ys)
   append(cons(x0, x1), ys)
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))


----------------------------------------

(32)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF2(true, n, x, xs) -> FILTERHIGH(n, xs)
   FILTERHIGH(n, cons(x, xs)) -> IF2(ge(x, n), n, x, xs)
   IF2(false, n, x, xs) -> FILTERHIGH(n, xs)

The TRS R consists of the following rules:

   ge(x, 0) -> true
   ge(0, s(x)) -> false
   ge(s(x), s(y)) -> ge(x, y)

The set Q consists of the following terms:

   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(33) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*FILTERHIGH(n, cons(x, xs)) -> IF2(ge(x, n), n, x, xs)
The graph contains the following edges 1 >= 2, 2 > 3, 2 > 4


*IF2(true, n, x, xs) -> FILTERHIGH(n, xs)
The graph contains the following edges 2 >= 1, 4 >= 2


*IF2(false, n, x, xs) -> FILTERHIGH(n, xs)
The graph contains the following edges 2 >= 1, 4 >= 2


----------------------------------------

(34)
YES

----------------------------------------

(35)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF1(true, n, x, xs) -> FILTERLOW(n, xs)
   FILTERLOW(n, cons(x, xs)) -> IF1(ge(n, x), n, x, xs)
   IF1(false, n, x, xs) -> FILTERLOW(n, xs)

The TRS R consists of the following rules:

   qsort(nil) -> nil
   qsort(cons(x, xs)) -> append(qsort(filterlow(last(cons(x, xs)), cons(x, xs))), cons(last(cons(x, xs)), qsort(filterhigh(last(cons(x, xs)), cons(x, xs)))))
   filterlow(n, nil) -> nil
   filterlow(n, cons(x, xs)) -> if1(ge(n, x), n, x, xs)
   if1(true, n, x, xs) -> filterlow(n, xs)
   if1(false, n, x, xs) -> cons(x, filterlow(n, xs))
   filterhigh(n, nil) -> nil
   filterhigh(n, cons(x, xs)) -> if2(ge(x, n), n, x, xs)
   if2(true, n, x, xs) -> filterhigh(n, xs)
   if2(false, n, x, xs) -> cons(x, filterhigh(n, xs))
   ge(x, 0) -> true
   ge(0, s(x)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   append(nil, ys) -> ys
   append(cons(x, xs), ys) -> cons(x, append(xs, ys))
   last(nil) -> 0
   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))

The set Q consists of the following terms:

   qsort(nil)
   qsort(cons(x0, x1))
   filterlow(x0, nil)
   filterlow(x0, cons(x1, x2))
   if1(true, x0, x1, x2)
   if1(false, x0, x1, x2)
   filterhigh(x0, nil)
   filterhigh(x0, cons(x1, x2))
   if2(true, x0, x1, x2)
   if2(false, x0, x1, x2)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   append(nil, ys)
   append(cons(x0, x1), ys)
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(36) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(37)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF1(true, n, x, xs) -> FILTERLOW(n, xs)
   FILTERLOW(n, cons(x, xs)) -> IF1(ge(n, x), n, x, xs)
   IF1(false, n, x, xs) -> FILTERLOW(n, xs)

The TRS R consists of the following rules:

   ge(x, 0) -> true
   ge(0, s(x)) -> false
   ge(s(x), s(y)) -> ge(x, y)

The set Q consists of the following terms:

   qsort(nil)
   qsort(cons(x0, x1))
   filterlow(x0, nil)
   filterlow(x0, cons(x1, x2))
   if1(true, x0, x1, x2)
   if1(false, x0, x1, x2)
   filterhigh(x0, nil)
   filterhigh(x0, cons(x1, x2))
   if2(true, x0, x1, x2)
   if2(false, x0, x1, x2)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   append(nil, ys)
   append(cons(x0, x1), ys)
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(38) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   qsort(nil)
   qsort(cons(x0, x1))
   filterlow(x0, nil)
   filterlow(x0, cons(x1, x2))
   if1(true, x0, x1, x2)
   if1(false, x0, x1, x2)
   filterhigh(x0, nil)
   filterhigh(x0, cons(x1, x2))
   if2(true, x0, x1, x2)
   if2(false, x0, x1, x2)
   append(nil, ys)
   append(cons(x0, x1), ys)
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))


----------------------------------------

(39)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF1(true, n, x, xs) -> FILTERLOW(n, xs)
   FILTERLOW(n, cons(x, xs)) -> IF1(ge(n, x), n, x, xs)
   IF1(false, n, x, xs) -> FILTERLOW(n, xs)

The TRS R consists of the following rules:

   ge(x, 0) -> true
   ge(0, s(x)) -> false
   ge(s(x), s(y)) -> ge(x, y)

The set Q consists of the following terms:

   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(40) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*FILTERLOW(n, cons(x, xs)) -> IF1(ge(n, x), n, x, xs)
The graph contains the following edges 1 >= 2, 2 > 3, 2 > 4


*IF1(true, n, x, xs) -> FILTERLOW(n, xs)
The graph contains the following edges 2 >= 1, 4 >= 2


*IF1(false, n, x, xs) -> FILTERLOW(n, xs)
The graph contains the following edges 2 >= 1, 4 >= 2


----------------------------------------

(41)
YES

----------------------------------------

(42)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   QSORT(cons(x, xs)) -> QSORT(filterhigh(last(cons(x, xs)), cons(x, xs)))
   QSORT(cons(x, xs)) -> QSORT(filterlow(last(cons(x, xs)), cons(x, xs)))

The TRS R consists of the following rules:

   qsort(nil) -> nil
   qsort(cons(x, xs)) -> append(qsort(filterlow(last(cons(x, xs)), cons(x, xs))), cons(last(cons(x, xs)), qsort(filterhigh(last(cons(x, xs)), cons(x, xs)))))
   filterlow(n, nil) -> nil
   filterlow(n, cons(x, xs)) -> if1(ge(n, x), n, x, xs)
   if1(true, n, x, xs) -> filterlow(n, xs)
   if1(false, n, x, xs) -> cons(x, filterlow(n, xs))
   filterhigh(n, nil) -> nil
   filterhigh(n, cons(x, xs)) -> if2(ge(x, n), n, x, xs)
   if2(true, n, x, xs) -> filterhigh(n, xs)
   if2(false, n, x, xs) -> cons(x, filterhigh(n, xs))
   ge(x, 0) -> true
   ge(0, s(x)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   append(nil, ys) -> ys
   append(cons(x, xs), ys) -> cons(x, append(xs, ys))
   last(nil) -> 0
   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))

The set Q consists of the following terms:

   qsort(nil)
   qsort(cons(x0, x1))
   filterlow(x0, nil)
   filterlow(x0, cons(x1, x2))
   if1(true, x0, x1, x2)
   if1(false, x0, x1, x2)
   filterhigh(x0, nil)
   filterhigh(x0, cons(x1, x2))
   if2(true, x0, x1, x2)
   if2(false, x0, x1, x2)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   append(nil, ys)
   append(cons(x0, x1), ys)
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(43) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(44)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   QSORT(cons(x, xs)) -> QSORT(filterhigh(last(cons(x, xs)), cons(x, xs)))
   QSORT(cons(x, xs)) -> QSORT(filterlow(last(cons(x, xs)), cons(x, xs)))

The TRS R consists of the following rules:

   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))
   filterlow(n, cons(x, xs)) -> if1(ge(n, x), n, x, xs)
   if1(true, n, x, xs) -> filterlow(n, xs)
   ge(x, 0) -> true
   ge(0, s(x)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   if1(false, n, x, xs) -> cons(x, filterlow(n, xs))
   filterlow(n, nil) -> nil
   filterhigh(n, cons(x, xs)) -> if2(ge(x, n), n, x, xs)
   if2(true, n, x, xs) -> filterhigh(n, xs)
   if2(false, n, x, xs) -> cons(x, filterhigh(n, xs))
   filterhigh(n, nil) -> nil

The set Q consists of the following terms:

   qsort(nil)
   qsort(cons(x0, x1))
   filterlow(x0, nil)
   filterlow(x0, cons(x1, x2))
   if1(true, x0, x1, x2)
   if1(false, x0, x1, x2)
   filterhigh(x0, nil)
   filterhigh(x0, cons(x1, x2))
   if2(true, x0, x1, x2)
   if2(false, x0, x1, x2)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   append(nil, ys)
   append(cons(x0, x1), ys)
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(45) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   qsort(nil)
   qsort(cons(x0, x1))
   append(nil, ys)
   append(cons(x0, x1), ys)


----------------------------------------

(46)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   QSORT(cons(x, xs)) -> QSORT(filterhigh(last(cons(x, xs)), cons(x, xs)))
   QSORT(cons(x, xs)) -> QSORT(filterlow(last(cons(x, xs)), cons(x, xs)))

The TRS R consists of the following rules:

   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))
   filterlow(n, cons(x, xs)) -> if1(ge(n, x), n, x, xs)
   if1(true, n, x, xs) -> filterlow(n, xs)
   ge(x, 0) -> true
   ge(0, s(x)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   if1(false, n, x, xs) -> cons(x, filterlow(n, xs))
   filterlow(n, nil) -> nil
   filterhigh(n, cons(x, xs)) -> if2(ge(x, n), n, x, xs)
   if2(true, n, x, xs) -> filterhigh(n, xs)
   if2(false, n, x, xs) -> cons(x, filterhigh(n, xs))
   filterhigh(n, nil) -> nil

The set Q consists of the following terms:

   filterlow(x0, nil)
   filterlow(x0, cons(x1, x2))
   if1(true, x0, x1, x2)
   if1(false, x0, x1, x2)
   filterhigh(x0, nil)
   filterhigh(x0, cons(x1, x2))
   if2(true, x0, x1, x2)
   if2(false, x0, x1, x2)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(47) Induction-Processor (SOUND)

This DP could be deleted by the Induction-Processor:
QSORT(cons(x, xs)) -> QSORT(filterhigh(last(cons(x, xs)), cons(x, xs)))


This order was computed:
Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(QSORT(x_1)) = x_1
   POL(cons(x_1, x_2)) = 1 + 2*x_1 + x_2
   POL(false_renamed) = 0
   POL(filterhigh(x_1, x_2)) = x_2
   POL(filterlow(x_1, x_2)) = x_2
   POL(ge(x_1, x_2)) = 0
   POL(if1(x_1, x_2, x_3, x_4)) = 1 + 2*x_3 + x_4
   POL(if2(x_1, x_2, x_3, x_4)) = 1 + 2*x_3 + x_4
   POL(last(x_1)) = x_1
   POL(nil) = 0
   POL(s(x_1)) = 2 + 3*x_1
   POL(true_renamed) = 0

At least one of these decreasing rules is always used after the deleted DP:
if2(true_renamed, n3, x9, xs5) -> filterhigh(n3, xs5)


The following formula is valid:
z0:sort[a37].(~(z0=nil)->filterhigh'(last(z0), z0)=true)


The transformed set:
filterhigh'(n2, cons(x8, xs4)) -> if2'(ge(x8, n2), n2, x8, xs4)
if2'(true_renamed, n3, x9, xs5) -> true
if2'(false_renamed, n4, x10, xs6) -> filterhigh'(n4, xs6)
filterhigh'(n5, nil) -> false
last(cons(x'', nil)) -> x''
last(cons(x1, cons(y, xs''))) -> last(cons(y, xs''))
filterlow(n, cons(x2, xs1)) -> if1(ge(n, x2), n, x2, xs1)
if1(true_renamed, n', x3, xs2) -> filterlow(n', xs2)
ge(x4, 0) -> true_renamed
ge(0, s(x5)) -> false_renamed
ge(s(x6), s(y')) -> ge(x6, y')
if1(false_renamed, n'', x7, xs3) -> cons(x7, filterlow(n'', xs3))
filterlow(n1, nil) -> nil
filterhigh(n2, cons(x8, xs4)) -> if2(ge(x8, n2), n2, x8, xs4)
if2(true_renamed, n3, x9, xs5) -> filterhigh(n3, xs5)
if2(false_renamed, n4, x10, xs6) -> cons(x10, filterhigh(n4, xs6))
filterhigh(n5, nil) -> nil
last(nil) -> 0
equal_bool(true, false) -> false
equal_bool(false, true) -> false
equal_bool(true, true) -> true
equal_bool(false, false) -> true
and(true, x) -> x
and(false, x) -> false
or(true, x) -> true
or(false, x) -> x
not(false) -> true
not(true) -> false
isa_true(true) -> true
isa_true(false) -> false
isa_false(true) -> false
isa_false(false) -> true
equal_sort[a0](0, 0) -> true
equal_sort[a0](0, s(v26)) -> false
equal_sort[a0](s(v27), 0) -> false
equal_sort[a0](s(v27), s(v28)) -> equal_sort[a0](v27, v28)
equal_sort[a37](cons(v29, v30), cons(v31, v32)) -> and(equal_sort[a0](v29, v31), equal_sort[a37](v30, v32))
equal_sort[a37](cons(v29, v30), nil) -> false
equal_sort[a37](nil, cons(v33, v34)) -> false
equal_sort[a37](nil, nil) -> true
equal_sort[a48](true_renamed, true_renamed) -> true
equal_sort[a48](true_renamed, false_renamed) -> false
equal_sort[a48](false_renamed, true_renamed) -> false
equal_sort[a48](false_renamed, false_renamed) -> true
equal_sort[a63](witness_sort[a63], witness_sort[a63]) -> true


The proof given by the theorem prover:
The following output was given by the internal theorem prover:proof of internal
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Partial correctness of the following Program

   [x, v26, v27, v28, v29, v30, v31, v32, v33, v34, n3, x9, xs5, n4, x10, x8, xs4, n5, n2, x'', x1, y, xs'', x4, x5, x6, y', n1, n, x2, xs1, n', x3, n'', x7]
   equal_bool(true, false) -> false
   equal_bool(false, true) -> false
   equal_bool(true, true) -> true
   equal_bool(false, false) -> true
   true and x -> x
   false and x -> false
   true or x -> true
   false or x -> x
   not(false) -> true
   not(true) -> false
   isa_true(true) -> true
   isa_true(false) -> false
   isa_false(true) -> false
   isa_false(false) -> true
   equal_sort[a0](0, 0) -> true
   equal_sort[a0](0, s(v26)) -> false
   equal_sort[a0](s(v27), 0) -> false
   equal_sort[a0](s(v27), s(v28)) -> equal_sort[a0](v27, v28)
   equal_sort[a37](cons(v29, v30), cons(v31, v32)) -> equal_sort[a0](v29, v31) and equal_sort[a37](v30, v32)
   equal_sort[a37](cons(v29, v30), nil) -> false
   equal_sort[a37](nil, cons(v33, v34)) -> false
   equal_sort[a37](nil, nil) -> true
   equal_sort[a48](true_renamed, true_renamed) -> true
   equal_sort[a48](true_renamed, false_renamed) -> false
   equal_sort[a48](false_renamed, true_renamed) -> false
   equal_sort[a48](false_renamed, false_renamed) -> true
   equal_sort[a63](witness_sort[a63], witness_sort[a63]) -> true
   if2'(true_renamed, n3, x9, xs5) -> true
   if2'(false_renamed, n4, x10, cons(x8, xs4)) -> if2'(ge(x8, n4), n4, x8, xs4)
   if2'(false_renamed, n4, x10, nil) -> false
   filterhigh'(n5, nil) -> false
   equal_sort[a48](ge(x8, n2), true_renamed) -> true | filterhigh'(n2, cons(x8, xs4)) -> true
   equal_sort[a48](ge(x8, n2), true_renamed) -> false | filterhigh'(n2, cons(x8, xs4)) -> filterhigh'(n2, xs4)
   last(cons(x'', nil)) -> x''
   last(cons(x1, cons(y, xs''))) -> last(cons(y, xs''))
   last(nil) -> 0
   ge(x4, 0) -> true_renamed
   ge(0, s(x5)) -> false_renamed
   ge(s(x6), s(y')) -> ge(x6, y')
   filterlow(n1, nil) -> nil
   equal_sort[a48](ge(n, x2), true_renamed) -> true | filterlow(n, cons(x2, xs1)) -> filterlow(n, xs1)
   equal_sort[a48](ge(n, x2), true_renamed) -> false | filterlow(n, cons(x2, xs1)) -> cons(x2, filterlow(n, xs1))
   filterhigh(n5, nil) -> nil
   equal_sort[a48](ge(x8, n2), true_renamed) -> true | filterhigh(n2, cons(x8, xs4)) -> filterhigh(n2, xs4)
   equal_sort[a48](ge(x8, n2), true_renamed) -> false | filterhigh(n2, cons(x8, xs4)) -> cons(x8, filterhigh(n2, xs4))
   if1(true_renamed, n', x3, cons(x2, xs1)) -> if1(ge(n', x2), n', x2, xs1)
   if1(true_renamed, n', x3, nil) -> nil
   if1(false_renamed, n'', x7, cons(x2, xs1)) -> cons(x7, if1(ge(n'', x2), n'', x2, xs1))
   if1(false_renamed, n'', x7, nil) -> cons(x7, nil)
   if2(true_renamed, n3, x9, cons(x8, xs4)) -> if2(ge(x8, n3), n3, x8, xs4)
   if2(true_renamed, n3, x9, nil) -> nil
   if2(false_renamed, n4, x10, cons(x8, xs4)) -> cons(x10, if2(ge(x8, n4), n4, x8, xs4))
   if2(false_renamed, n4, x10, nil) -> cons(x10, nil)

using the following formula:
z0:sort[a37].(~(z0=nil)->filterhigh'(last(z0), z0)=true)

could be successfully shown:
(0) Formula
(1) Induction by algorithm [EQUIVALENT, 0 ms]
(2) AND
    (3) Formula
        (4) Symbolic evaluation [EQUIVALENT, 0 ms]
        (5) Formula
        (6) Induction by data structure [EQUIVALENT, 0 ms]
        (7) AND
            (8) Formula
                (9) Symbolic evaluation [EQUIVALENT, 0 ms]
                (10) YES
            (11) Formula
                (12) Conditional Evaluation [EQUIVALENT, 0 ms]
                (13) AND
                    (14) Formula
                        (15) Symbolic evaluation [EQUIVALENT, 0 ms]
                        (16) YES
                    (17) Formula
                        (18) Symbolic evaluation [EQUIVALENT, 0 ms]
                        (19) Formula
                        (20) Hypothesis Lifting [EQUIVALENT, 0 ms]
                        (21) Formula
                        (22) Symbolic evaluation under hypothesis [SOUND, 0 ms]
                        (23) Formula
                        (24) Hypothesis Lifting [EQUIVALENT, 0 ms]
                        (25) Formula
                        (26) Hypothesis Lifting [EQUIVALENT, 0 ms]
                        (27) Formula
                        (28) Conditional Evaluation [EQUIVALENT, 0 ms]
                        (29) AND
                            (30) Formula
                                (31) Symbolic evaluation under hypothesis [EQUIVALENT, 0 ms]
                                (32) YES
                            (33) Formula
                                (34) Symbolic evaluation [EQUIVALENT, 0 ms]
                                (35) YES
    (36) Formula
        (37) Symbolic evaluation [EQUIVALENT, 0 ms]
        (38) YES
    (39) Formula
        (40) Symbolic evaluation [EQUIVALENT, 0 ms]
        (41) Formula
        (42) Conditional Evaluation [EQUIVALENT, 0 ms]
        (43) AND
            (44) Formula
                (45) Symbolic evaluation [EQUIVALENT, 0 ms]
                (46) YES
            (47) Formula
                (48) Symbolic evaluation under hypothesis [EQUIVALENT, 0 ms]
                (49) YES


----------------------------------------

(0)
Obligation:
Formula:
z0:sort[a37].(~(z0=nil)->filterhigh'(last(z0), z0)=true)

There are no hypotheses.




----------------------------------------

(1) Induction by algorithm (EQUIVALENT)
Induction by algorithm last(z0) generates the following cases:

1. Base Case:
Formula:
x'':sort[a0].(~(cons(x'', nil)=nil)->filterhigh'(last(cons(x'', nil)), cons(x'', nil))=true)

There are no hypotheses.





2. Base Case:
Formula:
(~(nil=nil)->filterhigh'(last(nil), nil)=true)

There are no hypotheses.





1. Step Case:
Formula:
x1:sort[a0],y:sort[a0],xs'':sort[a37].(~(cons(x1, cons(y, xs''))=nil)->filterhigh'(last(cons(x1, cons(y, xs''))), cons(x1, cons(y, xs'')))=true)

Hypotheses:
y:sort[a0],xs'':sort[a37].filterhigh'(last(cons(y, xs'')), cons(y, xs''))=true






----------------------------------------

(2)
Complex Obligation (AND)

----------------------------------------

(3)
Obligation:
Formula:
x'':sort[a0].(~(cons(x'', nil)=nil)->filterhigh'(last(cons(x'', nil)), cons(x'', nil))=true)

There are no hypotheses.




----------------------------------------

(4) Symbolic evaluation (EQUIVALENT)
Could be shown by simple symbolic evaluation.
----------------------------------------

(5)
Obligation:
Formula:
x'':sort[a0].filterhigh'(x'', cons(x'', nil))=true

There are no hypotheses.




----------------------------------------

(6) Induction by data structure (EQUIVALENT)
Induction by data structure sort[a0] generates the following cases:



1. Base Case:
Formula:
filterhigh'(0, cons(0, nil))=true

There are no hypotheses.





1. Step Case:
Formula:
n:sort[a0].filterhigh'(s(n), cons(s(n), nil))=true

Hypotheses:
n:sort[a0].filterhigh'(n, cons(n, nil))=true






----------------------------------------

(7)
Complex Obligation (AND)

----------------------------------------

(8)
Obligation:
Formula:
filterhigh'(0, cons(0, nil))=true

There are no hypotheses.




----------------------------------------

(9) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(10)
YES

----------------------------------------

(11)
Obligation:
Formula:
n:sort[a0].filterhigh'(s(n), cons(s(n), nil))=true

Hypotheses:
n:sort[a0].filterhigh'(n, cons(n, nil))=true




----------------------------------------

(12) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
true=true

Hypotheses:
n:sort[a0].filterhigh'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a48](ge(s(n), s(n)), true_renamed)=true





Formula:
n:sort[a0].filterhigh'(s(n), nil)=true

Hypotheses:
n:sort[a0].filterhigh'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a48](ge(s(n), s(n)), true_renamed)=false






----------------------------------------

(13)
Complex Obligation (AND)

----------------------------------------

(14)
Obligation:
Formula:
true=true

Hypotheses:
n:sort[a0].filterhigh'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a48](ge(s(n), s(n)), true_renamed)=true




----------------------------------------

(15) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(16)
YES

----------------------------------------

(17)
Obligation:
Formula:
n:sort[a0].filterhigh'(s(n), nil)=true

Hypotheses:
n:sort[a0].filterhigh'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a48](ge(s(n), s(n)), true_renamed)=false




----------------------------------------

(18) Symbolic evaluation (EQUIVALENT)
Could be shown by simple symbolic evaluation.
----------------------------------------

(19)
Obligation:
Formula:
False

Hypotheses:
n:sort[a0].filterhigh'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a48](ge(s(n), s(n)), true_renamed)=false




----------------------------------------

(20) Hypothesis Lifting (EQUIVALENT)
Formula could be generalised by hypothesis lifting to the following new obligation:
Formula:
n:sort[a0].((filterhigh'(n, cons(n, nil))=true/\equal_sort[a48](ge(s(n), s(n)), true_renamed)=false)->False)

Hypotheses:
n:sort[a0].filterhigh'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a48](ge(s(n), s(n)), true_renamed)=false




----------------------------------------

(21)
Obligation:
Formula:
n:sort[a0].((filterhigh'(n, cons(n, nil))=true/\equal_sort[a48](ge(s(n), s(n)), true_renamed)=false)->False)

Hypotheses:
n:sort[a0].filterhigh'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a48](ge(s(n), s(n)), true_renamed)=false




----------------------------------------

(22) Symbolic evaluation under hypothesis (SOUND)
Could be reduced by symbolic evaluation under hypothesis to:
n:sort[a0].~(equal_sort[a48](ge(n, n), true_renamed)=false)

By using the following hypotheses:
n:sort[a0].filterhigh'(n, cons(n, nil))=true

----------------------------------------

(23)
Obligation:
Formula:
n:sort[a0].~(equal_sort[a48](ge(n, n), true_renamed)=false)

Hypotheses:
n:sort[a0].filterhigh'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a48](ge(s(n), s(n)), true_renamed)=false




----------------------------------------

(24) Hypothesis Lifting (EQUIVALENT)
Formula could be generalised by hypothesis lifting to the following new obligation:
Formula:
n:sort[a0].(equal_sort[a48](ge(n, n), true_renamed)=false->~(equal_sort[a48](ge(n, n), true_renamed)=false))

Hypotheses:
n:sort[a0].filterhigh'(n, cons(n, nil))=true




----------------------------------------

(25)
Obligation:
Formula:
n:sort[a0].(equal_sort[a48](ge(n, n), true_renamed)=false->~(equal_sort[a48](ge(n, n), true_renamed)=false))

Hypotheses:
n:sort[a0].filterhigh'(n, cons(n, nil))=true




----------------------------------------

(26) Hypothesis Lifting (EQUIVALENT)
Formula could be generalised by hypothesis lifting to the following new obligation:
Formula:
n:sort[a0].(filterhigh'(n, cons(n, nil))=true->(equal_sort[a48](ge(n, n), true_renamed)=false->~(equal_sort[a48](ge(n, n), true_renamed)=false)))

There are no hypotheses.




----------------------------------------

(27)
Obligation:
Formula:
n:sort[a0].(filterhigh'(n, cons(n, nil))=true->(equal_sort[a48](ge(n, n), true_renamed)=false->~(equal_sort[a48](ge(n, n), true_renamed)=false)))

There are no hypotheses.




----------------------------------------

(28) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
n:sort[a0].(true=true->(equal_sort[a48](ge(n, n), true_renamed)=false->~(equal_sort[a48](ge(n, n), true_renamed)=false)))

Hypotheses:
n:sort[a0].equal_sort[a48](ge(n, n), true_renamed)=true





Formula:
n:sort[a0].(filterhigh'(n, nil)=true->(equal_sort[a48](ge(n, n), true_renamed)=false->~(equal_sort[a48](ge(n, n), true_renamed)=false)))

Hypotheses:
n:sort[a0].equal_sort[a48](ge(n, n), true_renamed)=false






----------------------------------------

(29)
Complex Obligation (AND)

----------------------------------------

(30)
Obligation:
Formula:
n:sort[a0].(true=true->(equal_sort[a48](ge(n, n), true_renamed)=false->~(equal_sort[a48](ge(n, n), true_renamed)=false)))

Hypotheses:
n:sort[a0].equal_sort[a48](ge(n, n), true_renamed)=true




----------------------------------------

(31) Symbolic evaluation under hypothesis (EQUIVALENT)
Could be shown using symbolic evaluation under hypothesis, by using the following hypotheses:

n:sort[a0].equal_sort[a48](ge(n, n), true_renamed)=true

----------------------------------------

(32)
YES

----------------------------------------

(33)
Obligation:
Formula:
n:sort[a0].(filterhigh'(n, nil)=true->(equal_sort[a48](ge(n, n), true_renamed)=false->~(equal_sort[a48](ge(n, n), true_renamed)=false)))

Hypotheses:
n:sort[a0].equal_sort[a48](ge(n, n), true_renamed)=false




----------------------------------------

(34) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(35)
YES

----------------------------------------

(36)
Obligation:
Formula:
(~(nil=nil)->filterhigh'(last(nil), nil)=true)

There are no hypotheses.




----------------------------------------

(37) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(38)
YES

----------------------------------------

(39)
Obligation:
Formula:
x1:sort[a0],y:sort[a0],xs'':sort[a37].(~(cons(x1, cons(y, xs''))=nil)->filterhigh'(last(cons(x1, cons(y, xs''))), cons(x1, cons(y, xs'')))=true)

Hypotheses:
y:sort[a0],xs'':sort[a37].filterhigh'(last(cons(y, xs'')), cons(y, xs''))=true




----------------------------------------

(40) Symbolic evaluation (EQUIVALENT)
Could be shown by simple symbolic evaluation.
----------------------------------------

(41)
Obligation:
Formula:
y:sort[a0],xs'':sort[a37],x1:sort[a0].filterhigh'(last(cons(y, xs'')), cons(x1, cons(y, xs'')))=true

Hypotheses:
y:sort[a0],xs'':sort[a37].filterhigh'(last(cons(y, xs'')), cons(y, xs''))=true




----------------------------------------

(42) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
true=true

Hypotheses:
y:sort[a0],xs'':sort[a37].filterhigh'(last(cons(y, xs'')), cons(y, xs''))=true
x1:sort[a0],y:sort[a0],xs'':sort[a37].equal_sort[a48](ge(x1, last(cons(y, xs''))), true_renamed)=true





Formula:
y:sort[a0],xs'':sort[a37].filterhigh'(last(cons(y, xs'')), cons(y, xs''))=true

Hypotheses:
y:sort[a0],xs'':sort[a37].filterhigh'(last(cons(y, xs'')), cons(y, xs''))=true
x1:sort[a0],y:sort[a0],xs'':sort[a37].equal_sort[a48](ge(x1, last(cons(y, xs''))), true_renamed)=false






----------------------------------------

(43)
Complex Obligation (AND)

----------------------------------------

(44)
Obligation:
Formula:
true=true

Hypotheses:
y:sort[a0],xs'':sort[a37].filterhigh'(last(cons(y, xs'')), cons(y, xs''))=true
x1:sort[a0],y:sort[a0],xs'':sort[a37].equal_sort[a48](ge(x1, last(cons(y, xs''))), true_renamed)=true




----------------------------------------

(45) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(46)
YES

----------------------------------------

(47)
Obligation:
Formula:
y:sort[a0],xs'':sort[a37].filterhigh'(last(cons(y, xs'')), cons(y, xs''))=true

Hypotheses:
y:sort[a0],xs'':sort[a37].filterhigh'(last(cons(y, xs'')), cons(y, xs''))=true
x1:sort[a0],y:sort[a0],xs'':sort[a37].equal_sort[a48](ge(x1, last(cons(y, xs''))), true_renamed)=false




----------------------------------------

(48) Symbolic evaluation under hypothesis (EQUIVALENT)
Could be shown using symbolic evaluation under hypothesis, by using the following hypotheses:

y:sort[a0],xs'':sort[a37].filterhigh'(last(cons(y, xs'')), cons(y, xs''))=true

----------------------------------------

(49)
YES

----------------------------------------

(48)
Complex Obligation (AND)

----------------------------------------

(49)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   QSORT(cons(x, xs)) -> QSORT(filterlow(last(cons(x, xs)), cons(x, xs)))

The TRS R consists of the following rules:

   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))
   filterlow(n, cons(x, xs)) -> if1(ge(n, x), n, x, xs)
   if1(true, n, x, xs) -> filterlow(n, xs)
   ge(x, 0) -> true
   ge(0, s(x)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   if1(false, n, x, xs) -> cons(x, filterlow(n, xs))
   filterlow(n, nil) -> nil
   filterhigh(n, cons(x, xs)) -> if2(ge(x, n), n, x, xs)
   if2(true, n, x, xs) -> filterhigh(n, xs)
   if2(false, n, x, xs) -> cons(x, filterhigh(n, xs))
   filterhigh(n, nil) -> nil

The set Q consists of the following terms:

   filterlow(x0, nil)
   filterlow(x0, cons(x1, x2))
   if1(true, x0, x1, x2)
   if1(false, x0, x1, x2)
   filterhigh(x0, nil)
   filterhigh(x0, cons(x1, x2))
   if2(true, x0, x1, x2)
   if2(false, x0, x1, x2)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(50) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(51)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   QSORT(cons(x, xs)) -> QSORT(filterlow(last(cons(x, xs)), cons(x, xs)))

The TRS R consists of the following rules:

   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))
   if1(true, n, x, xs) -> filterlow(n, xs)
   filterlow(n, cons(x, xs)) -> if1(ge(n, x), n, x, xs)
   filterlow(n, nil) -> nil
   ge(x, 0) -> true
   ge(0, s(x)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   if1(false, n, x, xs) -> cons(x, filterlow(n, xs))

The set Q consists of the following terms:

   filterlow(x0, nil)
   filterlow(x0, cons(x1, x2))
   if1(true, x0, x1, x2)
   if1(false, x0, x1, x2)
   filterhigh(x0, nil)
   filterhigh(x0, cons(x1, x2))
   if2(true, x0, x1, x2)
   if2(false, x0, x1, x2)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(52) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   filterhigh(x0, nil)
   filterhigh(x0, cons(x1, x2))
   if2(true, x0, x1, x2)
   if2(false, x0, x1, x2)


----------------------------------------

(53)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   QSORT(cons(x, xs)) -> QSORT(filterlow(last(cons(x, xs)), cons(x, xs)))

The TRS R consists of the following rules:

   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))
   if1(true, n, x, xs) -> filterlow(n, xs)
   filterlow(n, cons(x, xs)) -> if1(ge(n, x), n, x, xs)
   filterlow(n, nil) -> nil
   ge(x, 0) -> true
   ge(0, s(x)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   if1(false, n, x, xs) -> cons(x, filterlow(n, xs))

The set Q consists of the following terms:

   filterlow(x0, nil)
   filterlow(x0, cons(x1, x2))
   if1(true, x0, x1, x2)
   if1(false, x0, x1, x2)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(54) Induction-Processor (SOUND)

This DP could be deleted by the Induction-Processor:
QSORT(cons(x, xs)) -> QSORT(filterlow(last(cons(x, xs)), cons(x, xs)))


This order was computed:
Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(QSORT(x_1)) = x_1
   POL(cons(x_1, x_2)) = 1 + x_1 + 2*x_2
   POL(false_renamed) = 0
   POL(filterlow(x_1, x_2)) = x_2
   POL(ge(x_1, x_2)) = 0
   POL(if1(x_1, x_2, x_3, x_4)) = 1 + x_3 + 2*x_4
   POL(last(x_1)) = x_1
   POL(nil) = 0
   POL(s(x_1)) = 2 + 3*x_1
   POL(true_renamed) = 0

At least one of these decreasing rules is always used after the deleted DP:
if1(true_renamed, n1, x6, xs2) -> filterlow(n1, xs2)


The following formula is valid:
z0:sort[a25].(~(z0=nil)->filterlow'(last(z0), z0)=true)


The transformed set:
filterlow'(n, cons(x1, xs'')) -> if1'(ge(n, x1), n, x1, xs'')
filterlow'(n', nil) -> false
if1'(false_renamed, n'', x5, xs1) -> filterlow'(n'', xs1)
if1'(true_renamed, n1, x6, xs2) -> true
last(cons(x', nil)) -> x'
last(cons(x'', cons(y, xs'))) -> last(cons(y, xs'))
filterlow(n, cons(x1, xs'')) -> if1(ge(n, x1), n, x1, xs'')
filterlow(n', nil) -> nil
ge(x2, 0) -> true_renamed
ge(0, s(x3)) -> false_renamed
ge(s(x4), s(y')) -> ge(x4, y')
if1(false_renamed, n'', x5, xs1) -> cons(x5, filterlow(n'', xs1))
if1(true_renamed, n1, x6, xs2) -> filterlow(n1, xs2)
last(nil) -> 0
equal_bool(true, false) -> false
equal_bool(false, true) -> false
equal_bool(true, true) -> true
equal_bool(false, false) -> true
and(true, x) -> x
and(false, x) -> false
or(true, x) -> true
or(false, x) -> x
not(false) -> true
not(true) -> false
isa_true(true) -> true
isa_true(false) -> false
isa_false(true) -> false
isa_false(false) -> true
equal_sort[a0](0, 0) -> true
equal_sort[a0](0, s(v18)) -> false
equal_sort[a0](s(v19), 0) -> false
equal_sort[a0](s(v19), s(v20)) -> equal_sort[a0](v19, v20)
equal_sort[a25](cons(v21, v22), cons(v23, v24)) -> and(equal_sort[a0](v21, v23), equal_sort[a25](v22, v24))
equal_sort[a25](cons(v21, v22), nil) -> false
equal_sort[a25](nil, cons(v25, v26)) -> false
equal_sort[a25](nil, nil) -> true
equal_sort[a36](true_renamed, true_renamed) -> true
equal_sort[a36](true_renamed, false_renamed) -> false
equal_sort[a36](false_renamed, true_renamed) -> false
equal_sort[a36](false_renamed, false_renamed) -> true
equal_sort[a43](witness_sort[a43], witness_sort[a43]) -> true


The proof given by the theorem prover:
The following output was given by the internal theorem prover:proof of internal
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Partial correctness of the following Program

   [x, v18, v19, v20, v21, v22, v23, v24, v25, v26, n', n, x1, xs'', n1, x6, xs2, n'', x5, x', x'', y, xs', x2, x3, x4, y']
   equal_bool(true, false) -> false
   equal_bool(false, true) -> false
   equal_bool(true, true) -> true
   equal_bool(false, false) -> true
   true and x -> x
   false and x -> false
   true or x -> true
   false or x -> x
   not(false) -> true
   not(true) -> false
   isa_true(true) -> true
   isa_true(false) -> false
   isa_false(true) -> false
   isa_false(false) -> true
   equal_sort[a0](0, 0) -> true
   equal_sort[a0](0, s(v18)) -> false
   equal_sort[a0](s(v19), 0) -> false
   equal_sort[a0](s(v19), s(v20)) -> equal_sort[a0](v19, v20)
   equal_sort[a25](cons(v21, v22), cons(v23, v24)) -> equal_sort[a0](v21, v23) and equal_sort[a25](v22, v24)
   equal_sort[a25](cons(v21, v22), nil) -> false
   equal_sort[a25](nil, cons(v25, v26)) -> false
   equal_sort[a25](nil, nil) -> true
   equal_sort[a36](true_renamed, true_renamed) -> true
   equal_sort[a36](true_renamed, false_renamed) -> false
   equal_sort[a36](false_renamed, true_renamed) -> false
   equal_sort[a36](false_renamed, false_renamed) -> true
   equal_sort[a43](witness_sort[a43], witness_sort[a43]) -> true
   filterlow'(n', nil) -> false
   equal_sort[a36](ge(n, x1), false_renamed) -> true | filterlow'(n, cons(x1, xs'')) -> filterlow'(n, xs'')
   equal_sort[a36](ge(n, x1), false_renamed) -> false | filterlow'(n, cons(x1, xs'')) -> true
   if1'(true_renamed, n1, x6, xs2) -> true
   if1'(false_renamed, n'', x5, cons(x1, xs'')) -> if1'(ge(n'', x1), n'', x1, xs'')
   if1'(false_renamed, n'', x5, nil) -> false
   last(cons(x', nil)) -> x'
   last(cons(x'', cons(y, xs'))) -> last(cons(y, xs'))
   last(nil) -> 0
   filterlow(n', nil) -> nil
   equal_sort[a36](ge(n, x1), false_renamed) -> true | filterlow(n, cons(x1, xs'')) -> cons(x1, filterlow(n, xs''))
   equal_sort[a36](ge(n, x1), false_renamed) -> false | filterlow(n, cons(x1, xs'')) -> filterlow(n, xs'')
   ge(x2, 0) -> true_renamed
   ge(0, s(x3)) -> false_renamed
   ge(s(x4), s(y')) -> ge(x4, y')
   if1(false_renamed, n'', x5, cons(x1, xs'')) -> cons(x5, if1(ge(n'', x1), n'', x1, xs''))
   if1(false_renamed, n'', x5, nil) -> cons(x5, nil)
   if1(true_renamed, n1, x6, cons(x1, xs'')) -> if1(ge(n1, x1), n1, x1, xs'')
   if1(true_renamed, n1, x6, nil) -> nil

using the following formula:
z0:sort[a25].(~(z0=nil)->filterlow'(last(z0), z0)=true)

could be successfully shown:
(0) Formula
(1) Induction by algorithm [EQUIVALENT, 0 ms]
(2) AND
    (3) Formula
        (4) Symbolic evaluation [EQUIVALENT, 0 ms]
        (5) Formula
        (6) Induction by data structure [EQUIVALENT, 0 ms]
        (7) AND
            (8) Formula
                (9) Symbolic evaluation [EQUIVALENT, 0 ms]
                (10) YES
            (11) Formula
                (12) Conditional Evaluation [EQUIVALENT, 0 ms]
                (13) AND
                    (14) Formula
                        (15) Symbolic evaluation [EQUIVALENT, 0 ms]
                        (16) Formula
                        (17) Hypothesis Lifting [EQUIVALENT, 0 ms]
                        (18) Formula
                        (19) Symbolic evaluation under hypothesis [SOUND, 0 ms]
                        (20) Formula
                        (21) Hypothesis Lifting [EQUIVALENT, 0 ms]
                        (22) Formula
                        (23) Hypothesis Lifting [EQUIVALENT, 0 ms]
                        (24) Formula
                        (25) Conditional Evaluation [EQUIVALENT, 0 ms]
                        (26) AND
                            (27) Formula
                                (28) Symbolic evaluation [EQUIVALENT, 0 ms]
                                (29) YES
                            (30) Formula
                                (31) Symbolic evaluation under hypothesis [EQUIVALENT, 0 ms]
                                (32) YES
                    (33) Formula
                        (34) Symbolic evaluation [EQUIVALENT, 0 ms]
                        (35) YES
    (36) Formula
        (37) Symbolic evaluation [EQUIVALENT, 0 ms]
        (38) YES
    (39) Formula
        (40) Symbolic evaluation [EQUIVALENT, 0 ms]
        (41) Formula
        (42) Conditional Evaluation [EQUIVALENT, 0 ms]
        (43) AND
            (44) Formula
                (45) Symbolic evaluation under hypothesis [EQUIVALENT, 0 ms]
                (46) YES
            (47) Formula
                (48) Symbolic evaluation [EQUIVALENT, 0 ms]
                (49) YES


----------------------------------------

(0)
Obligation:
Formula:
z0:sort[a25].(~(z0=nil)->filterlow'(last(z0), z0)=true)

There are no hypotheses.




----------------------------------------

(1) Induction by algorithm (EQUIVALENT)
Induction by algorithm last(z0) generates the following cases:

1. Base Case:
Formula:
x':sort[a0].(~(cons(x', nil)=nil)->filterlow'(last(cons(x', nil)), cons(x', nil))=true)

There are no hypotheses.





2. Base Case:
Formula:
(~(nil=nil)->filterlow'(last(nil), nil)=true)

There are no hypotheses.





1. Step Case:
Formula:
x'':sort[a0],y:sort[a0],xs':sort[a25].(~(cons(x'', cons(y, xs'))=nil)->filterlow'(last(cons(x'', cons(y, xs'))), cons(x'', cons(y, xs')))=true)

Hypotheses:
y:sort[a0],xs':sort[a25].filterlow'(last(cons(y, xs')), cons(y, xs'))=true






----------------------------------------

(2)
Complex Obligation (AND)

----------------------------------------

(3)
Obligation:
Formula:
x':sort[a0].(~(cons(x', nil)=nil)->filterlow'(last(cons(x', nil)), cons(x', nil))=true)

There are no hypotheses.




----------------------------------------

(4) Symbolic evaluation (EQUIVALENT)
Could be shown by simple symbolic evaluation.
----------------------------------------

(5)
Obligation:
Formula:
x':sort[a0].filterlow'(x', cons(x', nil))=true

There are no hypotheses.




----------------------------------------

(6) Induction by data structure (EQUIVALENT)
Induction by data structure sort[a0] generates the following cases:



1. Base Case:
Formula:
filterlow'(0, cons(0, nil))=true

There are no hypotheses.





1. Step Case:
Formula:
n:sort[a0].filterlow'(s(n), cons(s(n), nil))=true

Hypotheses:
n:sort[a0].filterlow'(n, cons(n, nil))=true






----------------------------------------

(7)
Complex Obligation (AND)

----------------------------------------

(8)
Obligation:
Formula:
filterlow'(0, cons(0, nil))=true

There are no hypotheses.




----------------------------------------

(9) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(10)
YES

----------------------------------------

(11)
Obligation:
Formula:
n:sort[a0].filterlow'(s(n), cons(s(n), nil))=true

Hypotheses:
n:sort[a0].filterlow'(n, cons(n, nil))=true




----------------------------------------

(12) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
n:sort[a0].filterlow'(s(n), nil)=true

Hypotheses:
n:sort[a0].filterlow'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a36](ge(s(n), s(n)), false_renamed)=true





Formula:
true=true

Hypotheses:
n:sort[a0].filterlow'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a36](ge(s(n), s(n)), false_renamed)=false






----------------------------------------

(13)
Complex Obligation (AND)

----------------------------------------

(14)
Obligation:
Formula:
n:sort[a0].filterlow'(s(n), nil)=true

Hypotheses:
n:sort[a0].filterlow'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a36](ge(s(n), s(n)), false_renamed)=true




----------------------------------------

(15) Symbolic evaluation (EQUIVALENT)
Could be shown by simple symbolic evaluation.
----------------------------------------

(16)
Obligation:
Formula:
False

Hypotheses:
n:sort[a0].filterlow'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a36](ge(s(n), s(n)), false_renamed)=true




----------------------------------------

(17) Hypothesis Lifting (EQUIVALENT)
Formula could be generalised by hypothesis lifting to the following new obligation:
Formula:
n:sort[a0].((filterlow'(n, cons(n, nil))=true/\equal_sort[a36](ge(s(n), s(n)), false_renamed)=true)->False)

Hypotheses:
n:sort[a0].filterlow'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a36](ge(s(n), s(n)), false_renamed)=true




----------------------------------------

(18)
Obligation:
Formula:
n:sort[a0].((filterlow'(n, cons(n, nil))=true/\equal_sort[a36](ge(s(n), s(n)), false_renamed)=true)->False)

Hypotheses:
n:sort[a0].filterlow'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a36](ge(s(n), s(n)), false_renamed)=true




----------------------------------------

(19) Symbolic evaluation under hypothesis (SOUND)
Could be reduced by symbolic evaluation under hypothesis to:
n:sort[a0].~(equal_sort[a36](ge(n, n), false_renamed)=true)

By using the following hypotheses:
n:sort[a0].filterlow'(n, cons(n, nil))=true

----------------------------------------

(20)
Obligation:
Formula:
n:sort[a0].~(equal_sort[a36](ge(n, n), false_renamed)=true)

Hypotheses:
n:sort[a0].filterlow'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a36](ge(s(n), s(n)), false_renamed)=true




----------------------------------------

(21) Hypothesis Lifting (EQUIVALENT)
Formula could be generalised by hypothesis lifting to the following new obligation:
Formula:
n:sort[a0].(equal_sort[a36](ge(n, n), false_renamed)=true->~(equal_sort[a36](ge(n, n), false_renamed)=true))

Hypotheses:
n:sort[a0].filterlow'(n, cons(n, nil))=true




----------------------------------------

(22)
Obligation:
Formula:
n:sort[a0].(equal_sort[a36](ge(n, n), false_renamed)=true->~(equal_sort[a36](ge(n, n), false_renamed)=true))

Hypotheses:
n:sort[a0].filterlow'(n, cons(n, nil))=true




----------------------------------------

(23) Hypothesis Lifting (EQUIVALENT)
Formula could be generalised by hypothesis lifting to the following new obligation:
Formula:
n:sort[a0].(filterlow'(n, cons(n, nil))=true->(equal_sort[a36](ge(n, n), false_renamed)=true->~(equal_sort[a36](ge(n, n), false_renamed)=true)))

There are no hypotheses.




----------------------------------------

(24)
Obligation:
Formula:
n:sort[a0].(filterlow'(n, cons(n, nil))=true->(equal_sort[a36](ge(n, n), false_renamed)=true->~(equal_sort[a36](ge(n, n), false_renamed)=true)))

There are no hypotheses.




----------------------------------------

(25) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
n:sort[a0].(filterlow'(n, nil)=true->(equal_sort[a36](ge(n, n), false_renamed)=true->~(equal_sort[a36](ge(n, n), false_renamed)=true)))

Hypotheses:
n:sort[a0].equal_sort[a36](ge(n, n), false_renamed)=true





Formula:
n:sort[a0].(true=true->(equal_sort[a36](ge(n, n), false_renamed)=true->~(equal_sort[a36](ge(n, n), false_renamed)=true)))

Hypotheses:
n:sort[a0].equal_sort[a36](ge(n, n), false_renamed)=false






----------------------------------------

(26)
Complex Obligation (AND)

----------------------------------------

(27)
Obligation:
Formula:
n:sort[a0].(filterlow'(n, nil)=true->(equal_sort[a36](ge(n, n), false_renamed)=true->~(equal_sort[a36](ge(n, n), false_renamed)=true)))

Hypotheses:
n:sort[a0].equal_sort[a36](ge(n, n), false_renamed)=true




----------------------------------------

(28) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(29)
YES

----------------------------------------

(30)
Obligation:
Formula:
n:sort[a0].(true=true->(equal_sort[a36](ge(n, n), false_renamed)=true->~(equal_sort[a36](ge(n, n), false_renamed)=true)))

Hypotheses:
n:sort[a0].equal_sort[a36](ge(n, n), false_renamed)=false




----------------------------------------

(31) Symbolic evaluation under hypothesis (EQUIVALENT)
Could be shown using symbolic evaluation under hypothesis, by using the following hypotheses:

n:sort[a0].equal_sort[a36](ge(n, n), false_renamed)=false

----------------------------------------

(32)
YES

----------------------------------------

(33)
Obligation:
Formula:
true=true

Hypotheses:
n:sort[a0].filterlow'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a36](ge(s(n), s(n)), false_renamed)=false




----------------------------------------

(34) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(35)
YES

----------------------------------------

(36)
Obligation:
Formula:
(~(nil=nil)->filterlow'(last(nil), nil)=true)

There are no hypotheses.




----------------------------------------

(37) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(38)
YES

----------------------------------------

(39)
Obligation:
Formula:
x'':sort[a0],y:sort[a0],xs':sort[a25].(~(cons(x'', cons(y, xs'))=nil)->filterlow'(last(cons(x'', cons(y, xs'))), cons(x'', cons(y, xs')))=true)

Hypotheses:
y:sort[a0],xs':sort[a25].filterlow'(last(cons(y, xs')), cons(y, xs'))=true




----------------------------------------

(40) Symbolic evaluation (EQUIVALENT)
Could be shown by simple symbolic evaluation.
----------------------------------------

(41)
Obligation:
Formula:
y:sort[a0],xs':sort[a25],x'':sort[a0].filterlow'(last(cons(y, xs')), cons(x'', cons(y, xs')))=true

Hypotheses:
y:sort[a0],xs':sort[a25].filterlow'(last(cons(y, xs')), cons(y, xs'))=true




----------------------------------------

(42) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
y:sort[a0],xs':sort[a25].filterlow'(last(cons(y, xs')), cons(y, xs'))=true

Hypotheses:
y:sort[a0],xs':sort[a25].filterlow'(last(cons(y, xs')), cons(y, xs'))=true
y:sort[a0],xs':sort[a25],x'':sort[a0].equal_sort[a36](ge(last(cons(y, xs')), x''), false_renamed)=true





Formula:
true=true

Hypotheses:
y:sort[a0],xs':sort[a25].filterlow'(last(cons(y, xs')), cons(y, xs'))=true
y:sort[a0],xs':sort[a25],x'':sort[a0].equal_sort[a36](ge(last(cons(y, xs')), x''), false_renamed)=false






----------------------------------------

(43)
Complex Obligation (AND)

----------------------------------------

(44)
Obligation:
Formula:
y:sort[a0],xs':sort[a25].filterlow'(last(cons(y, xs')), cons(y, xs'))=true

Hypotheses:
y:sort[a0],xs':sort[a25].filterlow'(last(cons(y, xs')), cons(y, xs'))=true
y:sort[a0],xs':sort[a25],x'':sort[a0].equal_sort[a36](ge(last(cons(y, xs')), x''), false_renamed)=true




----------------------------------------

(45) Symbolic evaluation under hypothesis (EQUIVALENT)
Could be shown using symbolic evaluation under hypothesis, by using the following hypotheses:

y:sort[a0],xs':sort[a25].filterlow'(last(cons(y, xs')), cons(y, xs'))=true

----------------------------------------

(46)
YES

----------------------------------------

(47)
Obligation:
Formula:
true=true

Hypotheses:
y:sort[a0],xs':sort[a25].filterlow'(last(cons(y, xs')), cons(y, xs'))=true
y:sort[a0],xs':sort[a25],x'':sort[a0].equal_sort[a36](ge(last(cons(y, xs')), x''), false_renamed)=false




----------------------------------------

(48) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(49)
YES

----------------------------------------

(55)
Complex Obligation (AND)

----------------------------------------

(56)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))
   if1(true, n, x, xs) -> filterlow(n, xs)
   filterlow(n, cons(x, xs)) -> if1(ge(n, x), n, x, xs)
   filterlow(n, nil) -> nil
   ge(x, 0) -> true
   ge(0, s(x)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   if1(false, n, x, xs) -> cons(x, filterlow(n, xs))

The set Q consists of the following terms:

   filterlow(x0, nil)
   filterlow(x0, cons(x1, x2))
   if1(true, x0, x1, x2)
   if1(false, x0, x1, x2)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(57) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
----------------------------------------

(58)
YES

----------------------------------------

(59)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   filterlow'(n, cons(x1, xs'')) -> if1'(ge(n, x1), n, x1, xs'')
   filterlow'(n', nil) -> false
   if1'(false_renamed, n'', x5, xs1) -> filterlow'(n'', xs1)
   if1'(true_renamed, n1, x6, xs2) -> true
   last(cons(x', nil)) -> x'
   last(cons(x'', cons(y, xs'))) -> last(cons(y, xs'))
   filterlow(n, cons(x1, xs'')) -> if1(ge(n, x1), n, x1, xs'')
   filterlow(n', nil) -> nil
   ge(x2, 0) -> true_renamed
   ge(0, s(x3)) -> false_renamed
   ge(s(x4), s(y')) -> ge(x4, y')
   if1(false_renamed, n'', x5, xs1) -> cons(x5, filterlow(n'', xs1))
   if1(true_renamed, n1, x6, xs2) -> filterlow(n1, xs2)
   last(nil) -> 0
   equal_bool(true, false) -> false
   equal_bool(false, true) -> false
   equal_bool(true, true) -> true
   equal_bool(false, false) -> true
   and(true, x) -> x
   and(false, x) -> false
   or(true, x) -> true
   or(false, x) -> x
   not(false) -> true
   not(true) -> false
   isa_true(true) -> true
   isa_true(false) -> false
   isa_false(true) -> false
   isa_false(false) -> true
   equal_sort[a0](0, 0) -> true
   equal_sort[a0](0, s(v18)) -> false
   equal_sort[a0](s(v19), 0) -> false
   equal_sort[a0](s(v19), s(v20)) -> equal_sort[a0](v19, v20)
   equal_sort[a25](cons(v21, v22), cons(v23, v24)) -> and(equal_sort[a0](v21, v23), equal_sort[a25](v22, v24))
   equal_sort[a25](cons(v21, v22), nil) -> false
   equal_sort[a25](nil, cons(v25, v26)) -> false
   equal_sort[a25](nil, nil) -> true
   equal_sort[a36](true_renamed, true_renamed) -> true
   equal_sort[a36](true_renamed, false_renamed) -> false
   equal_sort[a36](false_renamed, true_renamed) -> false
   equal_sort[a36](false_renamed, false_renamed) -> true
   equal_sort[a43](witness_sort[a43], witness_sort[a43]) -> true

Q is empty.

----------------------------------------

(60) QTRSRRRProof (EQUIVALENT)
Used ordering:
filterlow'/2(YES,YES)
cons/2(YES,YES)
if1'/4(YES,YES,YES,YES)
ge/2(YES,YES)
nil/0)
false/0)
false_renamed/0)
true_renamed/0)
true/0)
last/1)YES(
filterlow/2(YES,YES)
if1/4(YES,YES,YES,YES)
0/0)
s/1)YES(
equal_bool/2(YES,YES)
and/2(YES,YES)
or/2(YES,YES)
not/1(YES)
isa_true/1(YES)
isa_false/1(YES)
equal_sort[a0]/2(YES,YES)
equal_sort[a25]/2(YES,YES)
equal_sort[a36]/2(YES,YES)
equal_sort[a43]/2(YES,YES)
witness_sort[a43]/0)

Quasi precedence:
[nil, false_renamed, true_renamed, filterlow_2, if1_4, 0] > [filterlow'_2, if1'_4] > ge_2
[nil, false_renamed, true_renamed, filterlow_2, if1_4, 0] > cons_2 > and_2
[nil, false_renamed, true_renamed, filterlow_2, if1_4, 0] > [true, equal_bool_2] > false
or_2 > [true, equal_bool_2] > false
not_1 > [true, equal_bool_2] > false
isa_true_1 > [true, equal_bool_2] > false
isa_false_1 > [true, equal_bool_2] > false
[equal_sort[a0]_2, equal_sort[a25]_2] > [true, equal_bool_2] > false
[equal_sort[a0]_2, equal_sort[a25]_2] > and_2
equal_sort[a36]_2 > [true, equal_bool_2] > false
equal_sort[a43]_2 > [true, equal_bool_2] > false


Status:
filterlow'_2: [2,1]
cons_2: multiset status
if1'_4: [4,2,3,1]
ge_2: multiset status
nil: multiset status
false: multiset status
false_renamed: multiset status
true_renamed: multiset status
true: multiset status
filterlow_2: [2,1]
if1_4: [4,2,3,1]
0: multiset status
equal_bool_2: [2,1]
and_2: [1,2]
or_2: [2,1]
not_1: multiset status
isa_true_1: [1]
isa_false_1: [1]
equal_sort[a0]_2: [1,2]
equal_sort[a25]_2: [1,2]
equal_sort[a36]_2: multiset status
equal_sort[a43]_2: [1,2]
witness_sort[a43]: multiset status

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   filterlow'(n, cons(x1, xs'')) -> if1'(ge(n, x1), n, x1, xs'')
   filterlow'(n', nil) -> false
   if1'(false_renamed, n'', x5, xs1) -> filterlow'(n'', xs1)
   if1'(true_renamed, n1, x6, xs2) -> true
   last(cons(x', nil)) -> x'
   last(cons(x'', cons(y, xs'))) -> last(cons(y, xs'))
   filterlow(n, cons(x1, xs'')) -> if1(ge(n, x1), n, x1, xs'')
   filterlow(n', nil) -> nil
   ge(x2, 0) -> true_renamed
   ge(0, s(x3)) -> false_renamed
   if1(false_renamed, n'', x5, xs1) -> cons(x5, filterlow(n'', xs1))
   if1(true_renamed, n1, x6, xs2) -> filterlow(n1, xs2)
   equal_bool(true, false) -> false
   equal_bool(false, true) -> false
   equal_bool(true, true) -> true
   equal_bool(false, false) -> true
   and(true, x) -> x
   and(false, x) -> false
   or(true, x) -> true
   or(false, x) -> x
   not(false) -> true
   not(true) -> false
   isa_true(true) -> true
   isa_true(false) -> false
   isa_false(true) -> false
   isa_false(false) -> true
   equal_sort[a0](0, 0) -> true
   equal_sort[a0](0, s(v18)) -> false
   equal_sort[a0](s(v19), 0) -> false
   equal_sort[a25](cons(v21, v22), cons(v23, v24)) -> and(equal_sort[a0](v21, v23), equal_sort[a25](v22, v24))
   equal_sort[a25](cons(v21, v22), nil) -> false
   equal_sort[a25](nil, cons(v25, v26)) -> false
   equal_sort[a25](nil, nil) -> true
   equal_sort[a36](true_renamed, true_renamed) -> true
   equal_sort[a36](true_renamed, false_renamed) -> false
   equal_sort[a36](false_renamed, true_renamed) -> false
   equal_sort[a36](false_renamed, false_renamed) -> true
   equal_sort[a43](witness_sort[a43], witness_sort[a43]) -> true




----------------------------------------

(61)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   ge(s(x4), s(y')) -> ge(x4, y')
   last(nil) -> 0
   equal_sort[a0](s(v19), s(v20)) -> equal_sort[a0](v19, v20)

Q is empty.

----------------------------------------

(62) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(0) = 2
   POL(equal_sort[a0](x_1, x_2)) = 2*x_1 + x_2
   POL(ge(x_1, x_2)) = 2*x_1 + 2*x_2
   POL(last(x_1)) = 1 + 2*x_1
   POL(nil) = 1
   POL(s(x_1)) = 2 + x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   ge(s(x4), s(y')) -> ge(x4, y')
   last(nil) -> 0
   equal_sort[a0](s(v19), s(v20)) -> equal_sort[a0](v19, v20)




----------------------------------------

(63)
Obligation:
Q restricted rewrite system:
R is empty.
Q is empty.

----------------------------------------

(64) RisEmptyProof (EQUIVALENT)
The TRS R is empty. Hence, termination is trivially proven.
----------------------------------------

(65)
YES

----------------------------------------

(66)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   filterhigh'(n2, cons(x8, xs4)) -> if2'(ge(x8, n2), n2, x8, xs4)
   if2'(true_renamed, n3, x9, xs5) -> true
   if2'(false_renamed, n4, x10, xs6) -> filterhigh'(n4, xs6)
   filterhigh'(n5, nil) -> false
   last(cons(x'', nil)) -> x''
   last(cons(x1, cons(y, xs''))) -> last(cons(y, xs''))
   filterlow(n, cons(x2, xs1)) -> if1(ge(n, x2), n, x2, xs1)
   if1(true_renamed, n', x3, xs2) -> filterlow(n', xs2)
   ge(x4, 0) -> true_renamed
   ge(0, s(x5)) -> false_renamed
   ge(s(x6), s(y')) -> ge(x6, y')
   if1(false_renamed, n'', x7, xs3) -> cons(x7, filterlow(n'', xs3))
   filterlow(n1, nil) -> nil
   filterhigh(n2, cons(x8, xs4)) -> if2(ge(x8, n2), n2, x8, xs4)
   if2(true_renamed, n3, x9, xs5) -> filterhigh(n3, xs5)
   if2(false_renamed, n4, x10, xs6) -> cons(x10, filterhigh(n4, xs6))
   filterhigh(n5, nil) -> nil
   last(nil) -> 0
   equal_bool(true, false) -> false
   equal_bool(false, true) -> false
   equal_bool(true, true) -> true
   equal_bool(false, false) -> true
   and(true, x) -> x
   and(false, x) -> false
   or(true, x) -> true
   or(false, x) -> x
   not(false) -> true
   not(true) -> false
   isa_true(true) -> true
   isa_true(false) -> false
   isa_false(true) -> false
   isa_false(false) -> true
   equal_sort[a0](0, 0) -> true
   equal_sort[a0](0, s(v26)) -> false
   equal_sort[a0](s(v27), 0) -> false
   equal_sort[a0](s(v27), s(v28)) -> equal_sort[a0](v27, v28)
   equal_sort[a37](cons(v29, v30), cons(v31, v32)) -> and(equal_sort[a0](v29, v31), equal_sort[a37](v30, v32))
   equal_sort[a37](cons(v29, v30), nil) -> false
   equal_sort[a37](nil, cons(v33, v34)) -> false
   equal_sort[a37](nil, nil) -> true
   equal_sort[a48](true_renamed, true_renamed) -> true
   equal_sort[a48](true_renamed, false_renamed) -> false
   equal_sort[a48](false_renamed, true_renamed) -> false
   equal_sort[a48](false_renamed, false_renamed) -> true
   equal_sort[a63](witness_sort[a63], witness_sort[a63]) -> true

Q is empty.

----------------------------------------

(67) QTRSRRRProof (EQUIVALENT)
Used ordering:
filterhigh'/2(YES,YES)
cons/2(YES,YES)
if2'/4(YES,YES,YES,YES)
ge/2(YES,YES)
true_renamed/0)
true/0)
false_renamed/0)
nil/0)
false/0)
last/1)YES(
filterlow/2(YES,YES)
if1/4(YES,YES,YES,YES)
0/0)
s/1)YES(
filterhigh/2(YES,YES)
if2/4(YES,YES,YES,YES)
equal_bool/2(YES,YES)
and/2(YES,YES)
or/2(YES,YES)
not/1(YES)
isa_true/1)YES(
isa_false/1(YES)
equal_sort[a0]/2(YES,YES)
equal_sort[a37]/2(YES,YES)
equal_sort[a48]/2(YES,YES)
equal_sort[a63]/2(YES,YES)
witness_sort[a63]/0)

Quasi precedence:
[filterhigh'_2, if2'_4] > [true, nil, equal_bool_2, equal_sort[a0]_2] > [ge_2, true_renamed, false_renamed, 0] > false
[filterlow_2, if1_4] > cons_2 > [ge_2, true_renamed, false_renamed, 0] > false
[filterlow_2, if1_4] > cons_2 > and_2 > false
[filterlow_2, if1_4] > [true, nil, equal_bool_2, equal_sort[a0]_2] > [ge_2, true_renamed, false_renamed, 0] > false
[filterhigh_2, if2_4] > cons_2 > [ge_2, true_renamed, false_renamed, 0] > false
[filterhigh_2, if2_4] > cons_2 > and_2 > false
[filterhigh_2, if2_4] > [true, nil, equal_bool_2, equal_sort[a0]_2] > [ge_2, true_renamed, false_renamed, 0] > false
or_2 > [true, nil, equal_bool_2, equal_sort[a0]_2] > [ge_2, true_renamed, false_renamed, 0] > false
not_1 > [true, nil, equal_bool_2, equal_sort[a0]_2] > [ge_2, true_renamed, false_renamed, 0] > false
isa_false_1 > [true, nil, equal_bool_2, equal_sort[a0]_2] > [ge_2, true_renamed, false_renamed, 0] > false
equal_sort[a37]_2 > [true, nil, equal_bool_2, equal_sort[a0]_2] > [ge_2, true_renamed, false_renamed, 0] > false
equal_sort[a37]_2 > and_2 > false
equal_sort[a48]_2 > [true, nil, equal_bool_2, equal_sort[a0]_2] > [ge_2, true_renamed, false_renamed, 0] > false
equal_sort[a63]_2 > [true, nil, equal_bool_2, equal_sort[a0]_2] > [ge_2, true_renamed, false_renamed, 0] > false
witness_sort[a63] > [true, nil, equal_bool_2, equal_sort[a0]_2] > [ge_2, true_renamed, false_renamed, 0] > false


Status:
filterhigh'_2: [2,1]
cons_2: multiset status
if2'_4: [4,2,3,1]
ge_2: multiset status
true_renamed: multiset status
true: multiset status
false_renamed: multiset status
nil: multiset status
false: multiset status
filterlow_2: [2,1]
if1_4: [4,2,3,1]
0: multiset status
filterhigh_2: [2,1]
if2_4: [4,2,1,3]
equal_bool_2: [1,2]
and_2: [2,1]
or_2: [2,1]
not_1: [1]
isa_false_1: multiset status
equal_sort[a0]_2: [1,2]
equal_sort[a37]_2: [1,2]
equal_sort[a48]_2: [1,2]
equal_sort[a63]_2: [2,1]
witness_sort[a63]: multiset status

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   filterhigh'(n2, cons(x8, xs4)) -> if2'(ge(x8, n2), n2, x8, xs4)
   if2'(true_renamed, n3, x9, xs5) -> true
   if2'(false_renamed, n4, x10, xs6) -> filterhigh'(n4, xs6)
   filterhigh'(n5, nil) -> false
   last(cons(x'', nil)) -> x''
   last(cons(x1, cons(y, xs''))) -> last(cons(y, xs''))
   filterlow(n, cons(x2, xs1)) -> if1(ge(n, x2), n, x2, xs1)
   if1(true_renamed, n', x3, xs2) -> filterlow(n', xs2)
   ge(x4, 0) -> true_renamed
   ge(0, s(x5)) -> false_renamed
   if1(false_renamed, n'', x7, xs3) -> cons(x7, filterlow(n'', xs3))
   filterlow(n1, nil) -> nil
   filterhigh(n2, cons(x8, xs4)) -> if2(ge(x8, n2), n2, x8, xs4)
   if2(true_renamed, n3, x9, xs5) -> filterhigh(n3, xs5)
   if2(false_renamed, n4, x10, xs6) -> cons(x10, filterhigh(n4, xs6))
   filterhigh(n5, nil) -> nil
   last(nil) -> 0
   equal_bool(true, false) -> false
   equal_bool(false, true) -> false
   equal_bool(true, true) -> true
   equal_bool(false, false) -> true
   and(true, x) -> x
   and(false, x) -> false
   or(true, x) -> true
   or(false, x) -> x
   not(false) -> true
   not(true) -> false
   isa_false(true) -> false
   isa_false(false) -> true
   equal_sort[a0](0, 0) -> true
   equal_sort[a0](0, s(v26)) -> false
   equal_sort[a0](s(v27), 0) -> false
   equal_sort[a37](cons(v29, v30), cons(v31, v32)) -> and(equal_sort[a0](v29, v31), equal_sort[a37](v30, v32))
   equal_sort[a37](cons(v29, v30), nil) -> false
   equal_sort[a37](nil, cons(v33, v34)) -> false
   equal_sort[a37](nil, nil) -> true
   equal_sort[a48](true_renamed, true_renamed) -> true
   equal_sort[a48](true_renamed, false_renamed) -> false
   equal_sort[a48](false_renamed, true_renamed) -> false
   equal_sort[a48](false_renamed, false_renamed) -> true
   equal_sort[a63](witness_sort[a63], witness_sort[a63]) -> true




----------------------------------------

(68)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   ge(s(x6), s(y')) -> ge(x6, y')
   isa_true(true) -> true
   isa_true(false) -> false
   equal_sort[a0](s(v27), s(v28)) -> equal_sort[a0](v27, v28)

Q is empty.

----------------------------------------

(69) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(equal_sort[a0](x_1, x_2)) = x_1 + x_2
   POL(false) = 0
   POL(ge(x_1, x_2)) = x_1 + x_2
   POL(isa_true(x_1)) = 2*x_1
   POL(s(x_1)) = 2 + 2*x_1
   POL(true) = 0
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   ge(s(x6), s(y')) -> ge(x6, y')
   equal_sort[a0](s(v27), s(v28)) -> equal_sort[a0](v27, v28)




----------------------------------------

(70)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   isa_true(true) -> true
   isa_true(false) -> false

Q is empty.

----------------------------------------

(71) QTRSRRRProof (EQUIVALENT)
Used ordering:
Knuth-Bendix order [KBO] with precedence:isa_true_1 > false > true

and weight map:

   true=1
   false=1
   isa_true_1=0

The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   isa_true(true) -> true
   isa_true(false) -> false




----------------------------------------

(72)
Obligation:
Q restricted rewrite system:
R is empty.
Q is empty.

----------------------------------------

(73) RisEmptyProof (EQUIVALENT)
The TRS R is empty. Hence, termination is trivially proven.
----------------------------------------

(74)
YES