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


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


YES
proof of /export/starexec/sandbox/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) Induction-Processor [SOUND, 59 ms]
        (34) AND
            (35) QDP
                (36) PisEmptyProof [EQUIVALENT, 0 ms]
                (37) YES
            (38) QTRS
                (39) QTRSRRRProof [EQUIVALENT, 64 ms]
                (40) QTRS
                (41) QTRSRRRProof [EQUIVALENT, 0 ms]
                (42) QTRS
                (43) RisEmptyProof [EQUIVALENT, 0 ms]
                (44) YES


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

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

   last(nil) -> 0
   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))
   del(x, nil) -> nil
   del(x, cons(y, xs)) -> if(eq(x, y), x, y, xs)
   if(true, x, y, xs) -> xs
   if(false, x, y, xs) -> cons(y, del(x, xs))
   eq(0, 0) -> true
   eq(0, s(y)) -> false
   eq(s(x), 0) -> false
   eq(s(x), s(y)) -> eq(x, y)
   reverse(nil) -> nil
   reverse(cons(x, xs)) -> cons(last(cons(x, xs)), reverse(del(last(cons(x, xs)), cons(x, 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:

   last(nil) -> 0
   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))
   del(x, nil) -> nil
   del(x, cons(y, xs)) -> if(eq(x, y), x, y, xs)
   if(true, x, y, xs) -> xs
   if(false, x, y, xs) -> cons(y, del(x, xs))
   eq(0, 0) -> true
   eq(0, s(y)) -> false
   eq(s(x), 0) -> false
   eq(s(x), s(y)) -> eq(x, y)
   reverse(nil) -> nil
   reverse(cons(x, xs)) -> cons(last(cons(x, xs)), reverse(del(last(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   del(x0, nil)
   del(x0, cons(x1, x2))
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   reverse(nil)
   reverse(cons(x0, x1))


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

(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:

   LAST(cons(x, cons(y, xs))) -> LAST(cons(y, xs))
   DEL(x, cons(y, xs)) -> IF(eq(x, y), x, y, xs)
   DEL(x, cons(y, xs)) -> EQ(x, y)
   IF(false, x, y, xs) -> DEL(x, xs)
   EQ(s(x), s(y)) -> EQ(x, y)
   REVERSE(cons(x, xs)) -> LAST(cons(x, xs))
   REVERSE(cons(x, xs)) -> REVERSE(del(last(cons(x, xs)), cons(x, xs)))
   REVERSE(cons(x, xs)) -> DEL(last(cons(x, xs)), cons(x, xs))

The TRS R consists of the following rules:

   last(nil) -> 0
   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))
   del(x, nil) -> nil
   del(x, cons(y, xs)) -> if(eq(x, y), x, y, xs)
   if(true, x, y, xs) -> xs
   if(false, x, y, xs) -> cons(y, del(x, xs))
   eq(0, 0) -> true
   eq(0, s(y)) -> false
   eq(s(x), 0) -> false
   eq(s(x), s(y)) -> eq(x, y)
   reverse(nil) -> nil
   reverse(cons(x, xs)) -> cons(last(cons(x, xs)), reverse(del(last(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   del(x0, nil)
   del(x0, cons(x1, x2))
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   reverse(nil)
   reverse(cons(x0, x1))

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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 3 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

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

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

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

The TRS R consists of the following rules:

   last(nil) -> 0
   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))
   del(x, nil) -> nil
   del(x, cons(y, xs)) -> if(eq(x, y), x, y, xs)
   if(true, x, y, xs) -> xs
   if(false, x, y, xs) -> cons(y, del(x, xs))
   eq(0, 0) -> true
   eq(0, s(y)) -> false
   eq(s(x), 0) -> false
   eq(s(x), s(y)) -> eq(x, y)
   reverse(nil) -> nil
   reverse(cons(x, xs)) -> cons(last(cons(x, xs)), reverse(del(last(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   del(x0, nil)
   del(x0, cons(x1, x2))
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   reverse(nil)
   reverse(cons(x0, x1))

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:

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

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

   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   del(x0, nil)
   del(x0, cons(x1, x2))
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   reverse(nil)
   reverse(cons(x0, x1))

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].

   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   del(x0, nil)
   del(x0, cons(x1, x2))
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   reverse(nil)
   reverse(cons(x0, x1))


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

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

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

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:
*EQ(s(x), s(y)) -> EQ(x, y)
The graph contains the following edges 1 > 1, 2 > 2


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

(13)
YES

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

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

   IF(false, x, y, xs) -> DEL(x, xs)
   DEL(x, cons(y, xs)) -> IF(eq(x, y), x, y, xs)

The TRS R consists of the following rules:

   last(nil) -> 0
   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))
   del(x, nil) -> nil
   del(x, cons(y, xs)) -> if(eq(x, y), x, y, xs)
   if(true, x, y, xs) -> xs
   if(false, x, y, xs) -> cons(y, del(x, xs))
   eq(0, 0) -> true
   eq(0, s(y)) -> false
   eq(s(x), 0) -> false
   eq(s(x), s(y)) -> eq(x, y)
   reverse(nil) -> nil
   reverse(cons(x, xs)) -> cons(last(cons(x, xs)), reverse(del(last(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   del(x0, nil)
   del(x0, cons(x1, x2))
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   reverse(nil)
   reverse(cons(x0, x1))

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:

   IF(false, x, y, xs) -> DEL(x, xs)
   DEL(x, cons(y, xs)) -> IF(eq(x, y), x, y, xs)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   del(x0, nil)
   del(x0, cons(x1, x2))
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   reverse(nil)
   reverse(cons(x0, x1))

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].

   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   del(x0, nil)
   del(x0, cons(x1, x2))
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   reverse(nil)
   reverse(cons(x0, x1))


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

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

   IF(false, x, y, xs) -> DEL(x, xs)
   DEL(x, cons(y, xs)) -> IF(eq(x, y), x, y, xs)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))

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:
*DEL(x, cons(y, xs)) -> IF(eq(x, y), x, y, xs)
The graph contains the following edges 1 >= 2, 2 > 3, 2 > 4


*IF(false, x, y, xs) -> DEL(x, xs)
The graph contains the following edges 2 >= 1, 4 >= 2


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

(20)
YES

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

(21)
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:

   last(nil) -> 0
   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))
   del(x, nil) -> nil
   del(x, cons(y, xs)) -> if(eq(x, y), x, y, xs)
   if(true, x, y, xs) -> xs
   if(false, x, y, xs) -> cons(y, del(x, xs))
   eq(0, 0) -> true
   eq(0, s(y)) -> false
   eq(s(x), 0) -> false
   eq(s(x), s(y)) -> eq(x, y)
   reverse(nil) -> nil
   reverse(cons(x, xs)) -> cons(last(cons(x, xs)), reverse(del(last(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   del(x0, nil)
   del(x0, cons(x1, x2))
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   reverse(nil)
   reverse(cons(x0, x1))

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:

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

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

   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   del(x0, nil)
   del(x0, cons(x1, x2))
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   reverse(nil)
   reverse(cons(x0, x1))

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].

   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   del(x0, nil)
   del(x0, cons(x1, x2))
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   reverse(nil)
   reverse(cons(x0, x1))


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

(25)
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.
----------------------------------------

(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:
*LAST(cons(x, cons(y, xs))) -> LAST(cons(y, xs))
The graph contains the following edges 1 > 1


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

(27)
YES

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

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

   REVERSE(cons(x, xs)) -> REVERSE(del(last(cons(x, xs)), cons(x, xs)))

The TRS R consists of the following rules:

   last(nil) -> 0
   last(cons(x, nil)) -> x
   last(cons(x, cons(y, xs))) -> last(cons(y, xs))
   del(x, nil) -> nil
   del(x, cons(y, xs)) -> if(eq(x, y), x, y, xs)
   if(true, x, y, xs) -> xs
   if(false, x, y, xs) -> cons(y, del(x, xs))
   eq(0, 0) -> true
   eq(0, s(y)) -> false
   eq(s(x), 0) -> false
   eq(s(x), s(y)) -> eq(x, y)
   reverse(nil) -> nil
   reverse(cons(x, xs)) -> cons(last(cons(x, xs)), reverse(del(last(cons(x, xs)), cons(x, xs))))

The set Q consists of the following terms:

   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   del(x0, nil)
   del(x0, cons(x1, x2))
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   reverse(nil)
   reverse(cons(x0, x1))

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:

   REVERSE(cons(x, xs)) -> REVERSE(del(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))
   del(x, cons(y, xs)) -> if(eq(x, y), x, y, xs)
   eq(0, 0) -> true
   eq(0, s(y)) -> false
   eq(s(x), 0) -> false
   eq(s(x), s(y)) -> eq(x, y)
   if(true, x, y, xs) -> xs
   if(false, x, y, xs) -> cons(y, del(x, xs))
   del(x, nil) -> nil

The set Q consists of the following terms:

   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   del(x0, nil)
   del(x0, cons(x1, x2))
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   reverse(nil)
   reverse(cons(x0, x1))

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].

   reverse(nil)
   reverse(cons(x0, x1))


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

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

   REVERSE(cons(x, xs)) -> REVERSE(del(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))
   del(x, cons(y, xs)) -> if(eq(x, y), x, y, xs)
   eq(0, 0) -> true
   eq(0, s(y)) -> false
   eq(s(x), 0) -> false
   eq(s(x), s(y)) -> eq(x, y)
   if(true, x, y, xs) -> xs
   if(false, x, y, xs) -> cons(y, del(x, xs))
   del(x, nil) -> nil

The set Q consists of the following terms:

   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   del(x0, nil)
   del(x0, cons(x1, x2))
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))

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

(33) Induction-Processor (SOUND)

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


This order was computed:
Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(REVERSE(x_1)) = x_1
   POL(cons(x_1, x_2)) = 1 + 2*x_1 + 2*x_2
   POL(del(x_1, x_2)) = x_2
   POL(eq(x_1, x_2)) = 0
   POL(false_renamed) = 0
   POL(if(x_1, x_2, x_3, x_4)) = 1 + 2*x_3 + 2*x_4
   POL(last(x_1)) = 2*x_1
   POL(nil) = 0
   POL(s(x_1)) = 3*x_1
   POL(true_renamed) = 0

At least one of these decreasing rules is always used after the deleted DP:
if(true_renamed, x4, y2, xs1) -> xs1


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


The transformed set:
del'(x1, cons(y', xs'')) -> if'(eq(x1, y'), x1, y', xs'')
if'(true_renamed, x4, y2, xs1) -> true
if'(false_renamed, x5, y3, xs2) -> del'(x5, xs2)
del'(x6, nil) -> false
last(cons(x', nil)) -> x'
last(cons(x'', cons(y, xs'))) -> last(cons(y, xs'))
del(x1, cons(y', xs'')) -> if(eq(x1, y'), x1, y', xs'')
eq(0, 0) -> true_renamed
eq(0, s(y'')) -> false_renamed
eq(s(x2), 0) -> false_renamed
eq(s(x3), s(y1)) -> eq(x3, y1)
if(true_renamed, x4, y2, xs1) -> xs1
if(false_renamed, x5, y3, xs2) -> cons(y3, del(x5, xs2))
del(x6, 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(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, x4, y2, xs1, x5, y3, y', xs'', x6, x1, x', x'', y, xs', y'', x2, x3, y1]
   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
   if'(true_renamed, x4, y2, xs1) -> true
   if'(false_renamed, x5, y3, cons(y', xs'')) -> if'(eq(x5, y'), x5, y', xs'')
   if'(false_renamed, x5, y3, nil) -> false
   del'(x6, nil) -> false
   equal_sort[a36](eq(x1, y'), true_renamed) -> true | del'(x1, cons(y', xs'')) -> true
   equal_sort[a36](eq(x1, y'), true_renamed) -> false | del'(x1, cons(y', xs'')) -> del'(x1, xs'')
   last(cons(x', nil)) -> x'
   last(cons(x'', cons(y, xs'))) -> last(cons(y, xs'))
   last(nil) -> 0
   eq(0, 0) -> true_renamed
   eq(0, s(y'')) -> false_renamed
   eq(s(x2), 0) -> false_renamed
   eq(s(x3), s(y1)) -> eq(x3, y1)
   if(true_renamed, x4, y2, xs1) -> xs1
   if(false_renamed, x5, y3, cons(y', xs'')) -> cons(y3, if(eq(x5, y'), x5, y', xs''))
   if(false_renamed, x5, y3, nil) -> cons(y3, nil)
   del(x6, nil) -> nil
   equal_sort[a36](eq(x1, y'), true_renamed) -> true | del(x1, cons(y', xs'')) -> xs''
   equal_sort[a36](eq(x1, y'), true_renamed) -> false | del(x1, cons(y', xs'')) -> cons(y', del(x1, xs''))

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

There are no hypotheses.





2. Base Case:
Formula:
(~(nil=nil)->del'(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)->del'(last(cons(x'', cons(y, xs'))), cons(x'', cons(y, xs')))=true)

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






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

(2)
Complex Obligation (AND)

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

(3)
Obligation:
Formula:
x':sort[a0].(~(cons(x', nil)=nil)->del'(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].del'(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:
del'(0, cons(0, nil))=true

There are no hypotheses.





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

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






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

(7)
Complex Obligation (AND)

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

(8)
Obligation:
Formula:
del'(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].del'(s(n), cons(s(n), nil))=true

Hypotheses:
n:sort[a0].del'(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].del'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a36](eq(s(n), s(n)), true_renamed)=true





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

Hypotheses:
n:sort[a0].del'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a36](eq(s(n), s(n)), true_renamed)=false






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

(13)
Complex Obligation (AND)

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

(14)
Obligation:
Formula:
true=true

Hypotheses:
n:sort[a0].del'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a36](eq(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].del'(s(n), nil)=true

Hypotheses:
n:sort[a0].del'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a36](eq(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].del'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a36](eq(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].((del'(n, cons(n, nil))=true/\equal_sort[a36](eq(s(n), s(n)), true_renamed)=false)->False)

Hypotheses:
n:sort[a0].del'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a36](eq(s(n), s(n)), true_renamed)=false




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

(21)
Obligation:
Formula:
n:sort[a0].((del'(n, cons(n, nil))=true/\equal_sort[a36](eq(s(n), s(n)), true_renamed)=false)->False)

Hypotheses:
n:sort[a0].del'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a36](eq(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[a36](eq(n, n), true_renamed)=false)

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

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

(23)
Obligation:
Formula:
n:sort[a0].~(equal_sort[a36](eq(n, n), true_renamed)=false)

Hypotheses:
n:sort[a0].del'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a36](eq(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[a36](eq(n, n), true_renamed)=false->~(equal_sort[a36](eq(n, n), true_renamed)=false))

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




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

(25)
Obligation:
Formula:
n:sort[a0].(equal_sort[a36](eq(n, n), true_renamed)=false->~(equal_sort[a36](eq(n, n), true_renamed)=false))

Hypotheses:
n:sort[a0].del'(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].(del'(n, cons(n, nil))=true->(equal_sort[a36](eq(n, n), true_renamed)=false->~(equal_sort[a36](eq(n, n), true_renamed)=false)))

There are no hypotheses.




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

(27)
Obligation:
Formula:
n:sort[a0].(del'(n, cons(n, nil))=true->(equal_sort[a36](eq(n, n), true_renamed)=false->~(equal_sort[a36](eq(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[a36](eq(n, n), true_renamed)=false->~(equal_sort[a36](eq(n, n), true_renamed)=false)))

Hypotheses:
n:sort[a0].equal_sort[a36](eq(n, n), true_renamed)=true





Formula:
n:sort[a0].(del'(n, nil)=true->(equal_sort[a36](eq(n, n), true_renamed)=false->~(equal_sort[a36](eq(n, n), true_renamed)=false)))

Hypotheses:
n:sort[a0].equal_sort[a36](eq(n, n), true_renamed)=false






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

(29)
Complex Obligation (AND)

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

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

Hypotheses:
n:sort[a0].equal_sort[a36](eq(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[a36](eq(n, n), true_renamed)=true

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

(32)
YES

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

(33)
Obligation:
Formula:
n:sort[a0].(del'(n, nil)=true->(equal_sort[a36](eq(n, n), true_renamed)=false->~(equal_sort[a36](eq(n, n), true_renamed)=false)))

Hypotheses:
n:sort[a0].equal_sort[a36](eq(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)->del'(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)->del'(last(cons(x'', cons(y, xs'))), cons(x'', cons(y, xs')))=true)

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

Hypotheses:
y:sort[a0],xs':sort[a25].del'(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[a25].del'(last(cons(y, xs')), cons(y, xs'))=true
y:sort[a0],xs':sort[a25],x'':sort[a0].equal_sort[a36](eq(last(cons(y, xs')), x''), true_renamed)=true





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

Hypotheses:
y:sort[a0],xs':sort[a25].del'(last(cons(y, xs')), cons(y, xs'))=true
y:sort[a0],xs':sort[a25],x'':sort[a0].equal_sort[a36](eq(last(cons(y, xs')), x''), true_renamed)=false






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

(43)
Complex Obligation (AND)

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

(44)
Obligation:
Formula:
true=true

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

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

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

(49)
YES

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

(34)
Complex Obligation (AND)

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

(35)
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))
   del(x, cons(y, xs)) -> if(eq(x, y), x, y, xs)
   eq(0, 0) -> true
   eq(0, s(y)) -> false
   eq(s(x), 0) -> false
   eq(s(x), s(y)) -> eq(x, y)
   if(true, x, y, xs) -> xs
   if(false, x, y, xs) -> cons(y, del(x, xs))
   del(x, nil) -> nil

The set Q consists of the following terms:

   last(nil)
   last(cons(x0, nil))
   last(cons(x0, cons(x1, x2)))
   del(x0, nil)
   del(x0, cons(x1, x2))
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)
   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))

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

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

(37)
YES

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

(38)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   del'(x1, cons(y', xs'')) -> if'(eq(x1, y'), x1, y', xs'')
   if'(true_renamed, x4, y2, xs1) -> true
   if'(false_renamed, x5, y3, xs2) -> del'(x5, xs2)
   del'(x6, nil) -> false
   last(cons(x', nil)) -> x'
   last(cons(x'', cons(y, xs'))) -> last(cons(y, xs'))
   del(x1, cons(y', xs'')) -> if(eq(x1, y'), x1, y', xs'')
   eq(0, 0) -> true_renamed
   eq(0, s(y'')) -> false_renamed
   eq(s(x2), 0) -> false_renamed
   eq(s(x3), s(y1)) -> eq(x3, y1)
   if(true_renamed, x4, y2, xs1) -> xs1
   if(false_renamed, x5, y3, xs2) -> cons(y3, del(x5, xs2))
   del(x6, 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(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.

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

(39) QTRSRRRProof (EQUIVALENT)
Used ordering:
del'/2(YES,YES)
cons/2(YES,YES)
if'/4(YES,YES,YES,YES)
eq/2(YES,YES)
true_renamed/0)
true/0)
false_renamed/0)
nil/0)
false/0)
last/1)YES(
del/2(YES,YES)
if/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:
[del'_2, if'_4, eq_2, false_renamed, nil, del_2, if_4, 0] > cons_2
[del'_2, if'_4, eq_2, false_renamed, nil, del_2, if_4, 0] > [true_renamed, false, not_1, isa_false_1, equal_sort[a0]_2] > true
or_2 > true
isa_true_1 > [true_renamed, false, not_1, isa_false_1, equal_sort[a0]_2] > true
equal_sort[a25]_2 > and_2 > [true_renamed, false, not_1, isa_false_1, equal_sort[a0]_2] > true
equal_sort[a43]_2 > true


Status:
del'_2: [1,2]
cons_2: multiset status
if'_4: [2,4,3,1]
eq_2: [1,2]
true_renamed: multiset status
true: multiset status
false_renamed: multiset status
nil: multiset status
false: multiset status
del_2: [1,2]
if_4: [2,4,1,3]
0: multiset status
equal_bool_2: multiset status
and_2: [2,1]
or_2: [2,1]
not_1: [1]
isa_true_1: [1]
isa_false_1: [1]
equal_sort[a0]_2: multiset status
equal_sort[a25]_2: multiset status
equal_sort[a36]_2: [2,1]
equal_sort[a43]_2: multiset status
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:

   del'(x1, cons(y', xs'')) -> if'(eq(x1, y'), x1, y', xs'')
   if'(true_renamed, x4, y2, xs1) -> true
   if'(false_renamed, x5, y3, xs2) -> del'(x5, xs2)
   del'(x6, nil) -> false
   last(cons(x', nil)) -> x'
   last(cons(x'', cons(y, xs'))) -> last(cons(y, xs'))
   del(x1, cons(y', xs'')) -> if(eq(x1, y'), x1, y', xs'')
   eq(0, 0) -> true_renamed
   eq(0, s(y'')) -> false_renamed
   eq(s(x2), 0) -> false_renamed
   if(true_renamed, x4, y2, xs1) -> xs1
   if(false_renamed, x5, y3, xs2) -> cons(y3, del(x5, xs2))
   del(x6, nil) -> nil
   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




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(40)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   eq(s(x3), s(y1)) -> eq(x3, y1)
   last(nil) -> 0
   equal_sort[a0](s(v19), s(v20)) -> equal_sort[a0](v19, v20)

Q is empty.

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

   POL(0) = 2
   POL(eq(x_1, x_2)) = 2*x_1 + 2*x_2
   POL(equal_sort[a0](x_1, x_2)) = 2*x_1 + 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:

   eq(s(x3), s(y1)) -> eq(x3, y1)
   last(nil) -> 0
   equal_sort[a0](s(v19), s(v20)) -> equal_sort[a0](v19, v20)




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(42)
Obligation:
Q restricted rewrite system:
R is empty.
Q is empty.

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(43) RisEmptyProof (EQUIVALENT)
The TRS R is empty. Hence, termination is trivially proven.
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(44)
YES