/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


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

(0) QTRS
(1) Overlay + Local Confluence [EQUIVALENT, 2 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, 2 ms]
        (18) QDP
        (19) QDPOrderProof [EQUIVALENT, 17 ms]
        (20) QDP
        (21) DependencyGraphProof [EQUIVALENT, 0 ms]
        (22) TRUE
    (23) QDP
        (24) UsableRulesProof [EQUIVALENT, 0 ms]
        (25) QDP
        (26) QReductionProof [EQUIVALENT, 0 ms]
        (27) QDP
        (28) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (29) YES
    (30) QDP
        (31) UsableRulesProof [EQUIVALENT, 0 ms]
        (32) QDP
        (33) QReductionProof [EQUIVALENT, 0 ms]
        (34) QDP
        (35) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (36) YES
    (37) QDP
        (38) UsableRulesProof [EQUIVALENT, 0 ms]
        (39) QDP
        (40) QReductionProof [EQUIVALENT, 1 ms]
        (41) QDP
        (42) QDPOrderProof [EQUIVALENT, 205 ms]
        (43) QDP
        (44) PisEmptyProof [EQUIVALENT, 0 ms]
        (45) YES


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

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

   eq(0, 0) -> true
   eq(0, s(m)) -> false
   eq(s(n), 0) -> false
   eq(s(n), s(m)) -> eq(n, m)
   le(0, m) -> true
   le(s(n), 0) -> false
   le(s(n), s(m)) -> le(n, m)
   min(cons(x, nil)) -> x
   min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
   if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
   if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
   replace(n, m, nil) -> nil
   replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
   if_replace(true, n, m, cons(k, x)) -> cons(m, x)
   if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
   empty(nil) -> true
   empty(cons(n, x)) -> false
   head(cons(n, x)) -> n
   tail(nil) -> nil
   tail(cons(n, x)) -> x
   sort(x) -> sortIter(x, nil)
   sortIter(x, y) -> if(empty(x), x, y, append(y, cons(min(x), nil)))
   if(true, x, y, z) -> y
   if(false, x, y, z) -> sortIter(replace(min(x), head(x), tail(x)), z)

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:

   eq(0, 0) -> true
   eq(0, s(m)) -> false
   eq(s(n), 0) -> false
   eq(s(n), s(m)) -> eq(n, m)
   le(0, m) -> true
   le(s(n), 0) -> false
   le(s(n), s(m)) -> le(n, m)
   min(cons(x, nil)) -> x
   min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
   if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
   if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
   replace(n, m, nil) -> nil
   replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
   if_replace(true, n, m, cons(k, x)) -> cons(m, x)
   if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
   empty(nil) -> true
   empty(cons(n, x)) -> false
   head(cons(n, x)) -> n
   tail(nil) -> nil
   tail(cons(n, x)) -> x
   sort(x) -> sortIter(x, nil)
   sortIter(x, y) -> if(empty(x), x, y, append(y, cons(min(x), nil)))
   if(true, x, y, z) -> y
   if(false, x, y, z) -> sortIter(replace(min(x), head(x), tail(x)), z)

The set Q consists of the following terms:

   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   min(cons(x0, nil))
   min(cons(x0, cons(x1, x2)))
   if_min(true, cons(x0, cons(x1, x2)))
   if_min(false, cons(x0, cons(x1, x2)))
   replace(x0, x1, nil)
   replace(x0, x1, cons(x2, x3))
   if_replace(true, x0, x1, cons(x2, x3))
   if_replace(false, x0, x1, cons(x2, x3))
   empty(nil)
   empty(cons(x0, x1))
   head(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   sort(x0)
   sortIter(x0, x1)
   if(true, x0, x1, x2)
   if(false, x0, 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:

   EQ(s(n), s(m)) -> EQ(n, m)
   LE(s(n), s(m)) -> LE(n, m)
   MIN(cons(n, cons(m, x))) -> IF_MIN(le(n, m), cons(n, cons(m, x)))
   MIN(cons(n, cons(m, x))) -> LE(n, m)
   IF_MIN(true, cons(n, cons(m, x))) -> MIN(cons(n, x))
   IF_MIN(false, cons(n, cons(m, x))) -> MIN(cons(m, x))
   REPLACE(n, m, cons(k, x)) -> IF_REPLACE(eq(n, k), n, m, cons(k, x))
   REPLACE(n, m, cons(k, x)) -> EQ(n, k)
   IF_REPLACE(false, n, m, cons(k, x)) -> REPLACE(n, m, x)
   SORT(x) -> SORTITER(x, nil)
   SORTITER(x, y) -> IF(empty(x), x, y, append(y, cons(min(x), nil)))
   SORTITER(x, y) -> EMPTY(x)
   SORTITER(x, y) -> MIN(x)
   IF(false, x, y, z) -> SORTITER(replace(min(x), head(x), tail(x)), z)
   IF(false, x, y, z) -> REPLACE(min(x), head(x), tail(x))
   IF(false, x, y, z) -> MIN(x)
   IF(false, x, y, z) -> HEAD(x)
   IF(false, x, y, z) -> TAIL(x)

The TRS R consists of the following rules:

   eq(0, 0) -> true
   eq(0, s(m)) -> false
   eq(s(n), 0) -> false
   eq(s(n), s(m)) -> eq(n, m)
   le(0, m) -> true
   le(s(n), 0) -> false
   le(s(n), s(m)) -> le(n, m)
   min(cons(x, nil)) -> x
   min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
   if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
   if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
   replace(n, m, nil) -> nil
   replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
   if_replace(true, n, m, cons(k, x)) -> cons(m, x)
   if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
   empty(nil) -> true
   empty(cons(n, x)) -> false
   head(cons(n, x)) -> n
   tail(nil) -> nil
   tail(cons(n, x)) -> x
   sort(x) -> sortIter(x, nil)
   sortIter(x, y) -> if(empty(x), x, y, append(y, cons(min(x), nil)))
   if(true, x, y, z) -> y
   if(false, x, y, z) -> sortIter(replace(min(x), head(x), tail(x)), z)

The set Q consists of the following terms:

   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   min(cons(x0, nil))
   min(cons(x0, cons(x1, x2)))
   if_min(true, cons(x0, cons(x1, x2)))
   if_min(false, cons(x0, cons(x1, x2)))
   replace(x0, x1, nil)
   replace(x0, x1, cons(x2, x3))
   if_replace(true, x0, x1, cons(x2, x3))
   if_replace(false, x0, x1, cons(x2, x3))
   empty(nil)
   empty(cons(x0, x1))
   head(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   sort(x0)
   sortIter(x0, x1)
   if(true, x0, x1, x2)
   if(false, x0, 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 5 SCCs with 9 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

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

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

   LE(s(n), s(m)) -> LE(n, m)

The TRS R consists of the following rules:

   eq(0, 0) -> true
   eq(0, s(m)) -> false
   eq(s(n), 0) -> false
   eq(s(n), s(m)) -> eq(n, m)
   le(0, m) -> true
   le(s(n), 0) -> false
   le(s(n), s(m)) -> le(n, m)
   min(cons(x, nil)) -> x
   min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
   if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
   if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
   replace(n, m, nil) -> nil
   replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
   if_replace(true, n, m, cons(k, x)) -> cons(m, x)
   if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
   empty(nil) -> true
   empty(cons(n, x)) -> false
   head(cons(n, x)) -> n
   tail(nil) -> nil
   tail(cons(n, x)) -> x
   sort(x) -> sortIter(x, nil)
   sortIter(x, y) -> if(empty(x), x, y, append(y, cons(min(x), nil)))
   if(true, x, y, z) -> y
   if(false, x, y, z) -> sortIter(replace(min(x), head(x), tail(x)), z)

The set Q consists of the following terms:

   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   min(cons(x0, nil))
   min(cons(x0, cons(x1, x2)))
   if_min(true, cons(x0, cons(x1, x2)))
   if_min(false, cons(x0, cons(x1, x2)))
   replace(x0, x1, nil)
   replace(x0, x1, cons(x2, x3))
   if_replace(true, x0, x1, cons(x2, x3))
   if_replace(false, x0, x1, cons(x2, x3))
   empty(nil)
   empty(cons(x0, x1))
   head(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   sort(x0)
   sortIter(x0, x1)
   if(true, x0, x1, x2)
   if(false, x0, 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:

   LE(s(n), s(m)) -> LE(n, m)

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

   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   min(cons(x0, nil))
   min(cons(x0, cons(x1, x2)))
   if_min(true, cons(x0, cons(x1, x2)))
   if_min(false, cons(x0, cons(x1, x2)))
   replace(x0, x1, nil)
   replace(x0, x1, cons(x2, x3))
   if_replace(true, x0, x1, cons(x2, x3))
   if_replace(false, x0, x1, cons(x2, x3))
   empty(nil)
   empty(cons(x0, x1))
   head(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   sort(x0)
   sortIter(x0, x1)
   if(true, x0, x1, x2)
   if(false, x0, 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].

   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   min(cons(x0, nil))
   min(cons(x0, cons(x1, x2)))
   if_min(true, cons(x0, cons(x1, x2)))
   if_min(false, cons(x0, cons(x1, x2)))
   replace(x0, x1, nil)
   replace(x0, x1, cons(x2, x3))
   if_replace(true, x0, x1, cons(x2, x3))
   if_replace(false, x0, x1, cons(x2, x3))
   empty(nil)
   empty(cons(x0, x1))
   head(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   sort(x0)
   sortIter(x0, x1)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)


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

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

   LE(s(n), s(m)) -> LE(n, m)

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:
*LE(s(n), s(m)) -> LE(n, m)
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:

   MIN(cons(n, cons(m, x))) -> IF_MIN(le(n, m), cons(n, cons(m, x)))
   IF_MIN(true, cons(n, cons(m, x))) -> MIN(cons(n, x))
   IF_MIN(false, cons(n, cons(m, x))) -> MIN(cons(m, x))

The TRS R consists of the following rules:

   eq(0, 0) -> true
   eq(0, s(m)) -> false
   eq(s(n), 0) -> false
   eq(s(n), s(m)) -> eq(n, m)
   le(0, m) -> true
   le(s(n), 0) -> false
   le(s(n), s(m)) -> le(n, m)
   min(cons(x, nil)) -> x
   min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
   if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
   if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
   replace(n, m, nil) -> nil
   replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
   if_replace(true, n, m, cons(k, x)) -> cons(m, x)
   if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
   empty(nil) -> true
   empty(cons(n, x)) -> false
   head(cons(n, x)) -> n
   tail(nil) -> nil
   tail(cons(n, x)) -> x
   sort(x) -> sortIter(x, nil)
   sortIter(x, y) -> if(empty(x), x, y, append(y, cons(min(x), nil)))
   if(true, x, y, z) -> y
   if(false, x, y, z) -> sortIter(replace(min(x), head(x), tail(x)), z)

The set Q consists of the following terms:

   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   min(cons(x0, nil))
   min(cons(x0, cons(x1, x2)))
   if_min(true, cons(x0, cons(x1, x2)))
   if_min(false, cons(x0, cons(x1, x2)))
   replace(x0, x1, nil)
   replace(x0, x1, cons(x2, x3))
   if_replace(true, x0, x1, cons(x2, x3))
   if_replace(false, x0, x1, cons(x2, x3))
   empty(nil)
   empty(cons(x0, x1))
   head(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   sort(x0)
   sortIter(x0, x1)
   if(true, x0, x1, x2)
   if(false, x0, 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:

   MIN(cons(n, cons(m, x))) -> IF_MIN(le(n, m), cons(n, cons(m, x)))
   IF_MIN(true, cons(n, cons(m, x))) -> MIN(cons(n, x))
   IF_MIN(false, cons(n, cons(m, x))) -> MIN(cons(m, x))

The TRS R consists of the following rules:

   le(0, m) -> true
   le(s(n), 0) -> false
   le(s(n), s(m)) -> le(n, m)

The set Q consists of the following terms:

   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   min(cons(x0, nil))
   min(cons(x0, cons(x1, x2)))
   if_min(true, cons(x0, cons(x1, x2)))
   if_min(false, cons(x0, cons(x1, x2)))
   replace(x0, x1, nil)
   replace(x0, x1, cons(x2, x3))
   if_replace(true, x0, x1, cons(x2, x3))
   if_replace(false, x0, x1, cons(x2, x3))
   empty(nil)
   empty(cons(x0, x1))
   head(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   sort(x0)
   sortIter(x0, x1)
   if(true, x0, x1, x2)
   if(false, x0, 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].

   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   min(cons(x0, nil))
   min(cons(x0, cons(x1, x2)))
   if_min(true, cons(x0, cons(x1, x2)))
   if_min(false, cons(x0, cons(x1, x2)))
   replace(x0, x1, nil)
   replace(x0, x1, cons(x2, x3))
   if_replace(true, x0, x1, cons(x2, x3))
   if_replace(false, x0, x1, cons(x2, x3))
   empty(nil)
   empty(cons(x0, x1))
   head(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   sort(x0)
   sortIter(x0, x1)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)


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

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

   MIN(cons(n, cons(m, x))) -> IF_MIN(le(n, m), cons(n, cons(m, x)))
   IF_MIN(true, cons(n, cons(m, x))) -> MIN(cons(n, x))
   IF_MIN(false, cons(n, cons(m, x))) -> MIN(cons(m, x))

The TRS R consists of the following rules:

   le(0, m) -> true
   le(s(n), 0) -> false
   le(s(n), s(m)) -> le(n, m)

The set Q consists of the following terms:

   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))

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

(19) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   IF_MIN(true, cons(n, cons(m, x))) -> MIN(cons(n, x))
   IF_MIN(false, cons(n, cons(m, x))) -> MIN(cons(m, x))
The remaining pairs can at least be oriented weakly.
Used ordering:  Combined order from the following AFS and order.
MIN(x1)  =  x1

cons(x1, x2)  =  cons(x2)

IF_MIN(x1, x2)  =  x2


Knuth-Bendix order [KBO] with precedence:trivial

and weight map:

   dummyConstant=1
   cons_1=1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none


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

(20)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MIN(cons(n, cons(m, x))) -> IF_MIN(le(n, m), cons(n, cons(m, x)))

The TRS R consists of the following rules:

   le(0, m) -> true
   le(s(n), 0) -> false
   le(s(n), s(m)) -> le(n, m)

The set Q consists of the following terms:

   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))

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

(21) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
----------------------------------------

(22)
TRUE

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

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

   EQ(s(n), s(m)) -> EQ(n, m)

The TRS R consists of the following rules:

   eq(0, 0) -> true
   eq(0, s(m)) -> false
   eq(s(n), 0) -> false
   eq(s(n), s(m)) -> eq(n, m)
   le(0, m) -> true
   le(s(n), 0) -> false
   le(s(n), s(m)) -> le(n, m)
   min(cons(x, nil)) -> x
   min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
   if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
   if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
   replace(n, m, nil) -> nil
   replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
   if_replace(true, n, m, cons(k, x)) -> cons(m, x)
   if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
   empty(nil) -> true
   empty(cons(n, x)) -> false
   head(cons(n, x)) -> n
   tail(nil) -> nil
   tail(cons(n, x)) -> x
   sort(x) -> sortIter(x, nil)
   sortIter(x, y) -> if(empty(x), x, y, append(y, cons(min(x), nil)))
   if(true, x, y, z) -> y
   if(false, x, y, z) -> sortIter(replace(min(x), head(x), tail(x)), z)

The set Q consists of the following terms:

   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   min(cons(x0, nil))
   min(cons(x0, cons(x1, x2)))
   if_min(true, cons(x0, cons(x1, x2)))
   if_min(false, cons(x0, cons(x1, x2)))
   replace(x0, x1, nil)
   replace(x0, x1, cons(x2, x3))
   if_replace(true, x0, x1, cons(x2, x3))
   if_replace(false, x0, x1, cons(x2, x3))
   empty(nil)
   empty(cons(x0, x1))
   head(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   sort(x0)
   sortIter(x0, x1)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)

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

(24) 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.
----------------------------------------

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

   EQ(s(n), s(m)) -> EQ(n, m)

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

   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   min(cons(x0, nil))
   min(cons(x0, cons(x1, x2)))
   if_min(true, cons(x0, cons(x1, x2)))
   if_min(false, cons(x0, cons(x1, x2)))
   replace(x0, x1, nil)
   replace(x0, x1, cons(x2, x3))
   if_replace(true, x0, x1, cons(x2, x3))
   if_replace(false, x0, x1, cons(x2, x3))
   empty(nil)
   empty(cons(x0, x1))
   head(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   sort(x0)
   sortIter(x0, x1)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)

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

(26) 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].

   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   min(cons(x0, nil))
   min(cons(x0, cons(x1, x2)))
   if_min(true, cons(x0, cons(x1, x2)))
   if_min(false, cons(x0, cons(x1, x2)))
   replace(x0, x1, nil)
   replace(x0, x1, cons(x2, x3))
   if_replace(true, x0, x1, cons(x2, x3))
   if_replace(false, x0, x1, cons(x2, x3))
   empty(nil)
   empty(cons(x0, x1))
   head(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   sort(x0)
   sortIter(x0, x1)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)


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

(27)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   EQ(s(n), s(m)) -> EQ(n, m)

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

(28) 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(n), s(m)) -> EQ(n, m)
The graph contains the following edges 1 > 1, 2 > 2


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

(29)
YES

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

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

   REPLACE(n, m, cons(k, x)) -> IF_REPLACE(eq(n, k), n, m, cons(k, x))
   IF_REPLACE(false, n, m, cons(k, x)) -> REPLACE(n, m, x)

The TRS R consists of the following rules:

   eq(0, 0) -> true
   eq(0, s(m)) -> false
   eq(s(n), 0) -> false
   eq(s(n), s(m)) -> eq(n, m)
   le(0, m) -> true
   le(s(n), 0) -> false
   le(s(n), s(m)) -> le(n, m)
   min(cons(x, nil)) -> x
   min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
   if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
   if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
   replace(n, m, nil) -> nil
   replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
   if_replace(true, n, m, cons(k, x)) -> cons(m, x)
   if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
   empty(nil) -> true
   empty(cons(n, x)) -> false
   head(cons(n, x)) -> n
   tail(nil) -> nil
   tail(cons(n, x)) -> x
   sort(x) -> sortIter(x, nil)
   sortIter(x, y) -> if(empty(x), x, y, append(y, cons(min(x), nil)))
   if(true, x, y, z) -> y
   if(false, x, y, z) -> sortIter(replace(min(x), head(x), tail(x)), z)

The set Q consists of the following terms:

   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   min(cons(x0, nil))
   min(cons(x0, cons(x1, x2)))
   if_min(true, cons(x0, cons(x1, x2)))
   if_min(false, cons(x0, cons(x1, x2)))
   replace(x0, x1, nil)
   replace(x0, x1, cons(x2, x3))
   if_replace(true, x0, x1, cons(x2, x3))
   if_replace(false, x0, x1, cons(x2, x3))
   empty(nil)
   empty(cons(x0, x1))
   head(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   sort(x0)
   sortIter(x0, x1)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)

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

(31) 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.
----------------------------------------

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

   REPLACE(n, m, cons(k, x)) -> IF_REPLACE(eq(n, k), n, m, cons(k, x))
   IF_REPLACE(false, n, m, cons(k, x)) -> REPLACE(n, m, x)

The TRS R consists of the following rules:

   eq(0, 0) -> true
   eq(0, s(m)) -> false
   eq(s(n), 0) -> false
   eq(s(n), s(m)) -> eq(n, m)

The set Q consists of the following terms:

   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   min(cons(x0, nil))
   min(cons(x0, cons(x1, x2)))
   if_min(true, cons(x0, cons(x1, x2)))
   if_min(false, cons(x0, cons(x1, x2)))
   replace(x0, x1, nil)
   replace(x0, x1, cons(x2, x3))
   if_replace(true, x0, x1, cons(x2, x3))
   if_replace(false, x0, x1, cons(x2, x3))
   empty(nil)
   empty(cons(x0, x1))
   head(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   sort(x0)
   sortIter(x0, x1)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)

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

(33) 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].

   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   min(cons(x0, nil))
   min(cons(x0, cons(x1, x2)))
   if_min(true, cons(x0, cons(x1, x2)))
   if_min(false, cons(x0, cons(x1, x2)))
   replace(x0, x1, nil)
   replace(x0, x1, cons(x2, x3))
   if_replace(true, x0, x1, cons(x2, x3))
   if_replace(false, x0, x1, cons(x2, x3))
   empty(nil)
   empty(cons(x0, x1))
   head(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   sort(x0)
   sortIter(x0, x1)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)


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

(34)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   REPLACE(n, m, cons(k, x)) -> IF_REPLACE(eq(n, k), n, m, cons(k, x))
   IF_REPLACE(false, n, m, cons(k, x)) -> REPLACE(n, m, x)

The TRS R consists of the following rules:

   eq(0, 0) -> true
   eq(0, s(m)) -> false
   eq(s(n), 0) -> false
   eq(s(n), s(m)) -> eq(n, m)

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.
----------------------------------------

(35) 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:
*IF_REPLACE(false, n, m, cons(k, x)) -> REPLACE(n, m, x)
The graph contains the following edges 2 >= 1, 3 >= 2, 4 > 3


*REPLACE(n, m, cons(k, x)) -> IF_REPLACE(eq(n, k), n, m, cons(k, x))
The graph contains the following edges 1 >= 2, 2 >= 3, 3 >= 4


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

(36)
YES

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

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

   IF(false, x, y, z) -> SORTITER(replace(min(x), head(x), tail(x)), z)
   SORTITER(x, y) -> IF(empty(x), x, y, append(y, cons(min(x), nil)))

The TRS R consists of the following rules:

   eq(0, 0) -> true
   eq(0, s(m)) -> false
   eq(s(n), 0) -> false
   eq(s(n), s(m)) -> eq(n, m)
   le(0, m) -> true
   le(s(n), 0) -> false
   le(s(n), s(m)) -> le(n, m)
   min(cons(x, nil)) -> x
   min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
   if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
   if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
   replace(n, m, nil) -> nil
   replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
   if_replace(true, n, m, cons(k, x)) -> cons(m, x)
   if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))
   empty(nil) -> true
   empty(cons(n, x)) -> false
   head(cons(n, x)) -> n
   tail(nil) -> nil
   tail(cons(n, x)) -> x
   sort(x) -> sortIter(x, nil)
   sortIter(x, y) -> if(empty(x), x, y, append(y, cons(min(x), nil)))
   if(true, x, y, z) -> y
   if(false, x, y, z) -> sortIter(replace(min(x), head(x), tail(x)), z)

The set Q consists of the following terms:

   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   min(cons(x0, nil))
   min(cons(x0, cons(x1, x2)))
   if_min(true, cons(x0, cons(x1, x2)))
   if_min(false, cons(x0, cons(x1, x2)))
   replace(x0, x1, nil)
   replace(x0, x1, cons(x2, x3))
   if_replace(true, x0, x1, cons(x2, x3))
   if_replace(false, x0, x1, cons(x2, x3))
   empty(nil)
   empty(cons(x0, x1))
   head(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   sort(x0)
   sortIter(x0, x1)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)

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

(38) 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.
----------------------------------------

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

   IF(false, x, y, z) -> SORTITER(replace(min(x), head(x), tail(x)), z)
   SORTITER(x, y) -> IF(empty(x), x, y, append(y, cons(min(x), nil)))

The TRS R consists of the following rules:

   empty(nil) -> true
   empty(cons(n, x)) -> false
   min(cons(x, nil)) -> x
   min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
   if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
   if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
   le(0, m) -> true
   le(s(n), 0) -> false
   le(s(n), s(m)) -> le(n, m)
   head(cons(n, x)) -> n
   tail(nil) -> nil
   tail(cons(n, x)) -> x
   replace(n, m, nil) -> nil
   replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
   eq(0, 0) -> true
   eq(0, s(m)) -> false
   eq(s(n), 0) -> false
   eq(s(n), s(m)) -> eq(n, m)
   if_replace(true, n, m, cons(k, x)) -> cons(m, x)
   if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))

The set Q consists of the following terms:

   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   min(cons(x0, nil))
   min(cons(x0, cons(x1, x2)))
   if_min(true, cons(x0, cons(x1, x2)))
   if_min(false, cons(x0, cons(x1, x2)))
   replace(x0, x1, nil)
   replace(x0, x1, cons(x2, x3))
   if_replace(true, x0, x1, cons(x2, x3))
   if_replace(false, x0, x1, cons(x2, x3))
   empty(nil)
   empty(cons(x0, x1))
   head(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))
   sort(x0)
   sortIter(x0, x1)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)

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

(40) 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].

   sort(x0)
   sortIter(x0, x1)
   if(true, x0, x1, x2)
   if(false, x0, x1, x2)


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

(41)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF(false, x, y, z) -> SORTITER(replace(min(x), head(x), tail(x)), z)
   SORTITER(x, y) -> IF(empty(x), x, y, append(y, cons(min(x), nil)))

The TRS R consists of the following rules:

   empty(nil) -> true
   empty(cons(n, x)) -> false
   min(cons(x, nil)) -> x
   min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
   if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
   if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
   le(0, m) -> true
   le(s(n), 0) -> false
   le(s(n), s(m)) -> le(n, m)
   head(cons(n, x)) -> n
   tail(nil) -> nil
   tail(cons(n, x)) -> x
   replace(n, m, nil) -> nil
   replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
   eq(0, 0) -> true
   eq(0, s(m)) -> false
   eq(s(n), 0) -> false
   eq(s(n), s(m)) -> eq(n, m)
   if_replace(true, n, m, cons(k, x)) -> cons(m, x)
   if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))

The set Q consists of the following terms:

   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   min(cons(x0, nil))
   min(cons(x0, cons(x1, x2)))
   if_min(true, cons(x0, cons(x1, x2)))
   if_min(false, cons(x0, cons(x1, x2)))
   replace(x0, x1, nil)
   replace(x0, x1, cons(x2, x3))
   if_replace(true, x0, x1, cons(x2, x3))
   if_replace(false, x0, x1, cons(x2, x3))
   empty(nil)
   empty(cons(x0, x1))
   head(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))

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

(42) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   IF(false, x, y, z) -> SORTITER(replace(min(x), head(x), tail(x)), z)
   SORTITER(x, y) -> IF(empty(x), x, y, append(y, cons(min(x), nil)))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO,RATPOLO]:

   POL(0) = 0
   POL(IF(x_1, x_2, x_3, x_4)) = [1/2]x_1 + [1/4]x_2
   POL(SORTITER(x_1, x_2)) = [1/4] + [1/2]x_1
   POL(append(x_1, x_2)) = 0
   POL(cons(x_1, x_2)) = [2] + [4]x_2
   POL(empty(x_1)) = [1/2]x_1
   POL(eq(x_1, x_2)) = 0
   POL(false) = [1]
   POL(head(x_1)) = 0
   POL(if_min(x_1, x_2)) = [1]
   POL(if_replace(x_1, x_2, x_3, x_4)) = [2]x_4
   POL(le(x_1, x_2)) = [4]
   POL(min(x_1)) = 0
   POL(nil) = 0
   POL(replace(x_1, x_2, x_3)) = [2]x_3
   POL(s(x_1)) = 0
   POL(tail(x_1)) = [1/4]x_1
   POL(true) = 0
The value of delta used in the strict ordering is 1/4.
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   tail(nil) -> nil
   tail(cons(n, x)) -> x
   replace(n, m, nil) -> nil
   replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
   empty(nil) -> true
   empty(cons(n, x)) -> false
   if_replace(true, n, m, cons(k, x)) -> cons(m, x)
   if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))


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

(43)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   empty(nil) -> true
   empty(cons(n, x)) -> false
   min(cons(x, nil)) -> x
   min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
   if_min(true, cons(n, cons(m, x))) -> min(cons(n, x))
   if_min(false, cons(n, cons(m, x))) -> min(cons(m, x))
   le(0, m) -> true
   le(s(n), 0) -> false
   le(s(n), s(m)) -> le(n, m)
   head(cons(n, x)) -> n
   tail(nil) -> nil
   tail(cons(n, x)) -> x
   replace(n, m, nil) -> nil
   replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x))
   eq(0, 0) -> true
   eq(0, s(m)) -> false
   eq(s(n), 0) -> false
   eq(s(n), s(m)) -> eq(n, m)
   if_replace(true, n, m, cons(k, x)) -> cons(m, x)
   if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x))

The set Q consists of the following terms:

   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   min(cons(x0, nil))
   min(cons(x0, cons(x1, x2)))
   if_min(true, cons(x0, cons(x1, x2)))
   if_min(false, cons(x0, cons(x1, x2)))
   replace(x0, x1, nil)
   replace(x0, x1, cons(x2, x3))
   if_replace(true, x0, x1, cons(x2, x3))
   if_replace(false, x0, x1, cons(x2, x3))
   empty(nil)
   empty(cons(x0, x1))
   head(cons(x0, x1))
   tail(nil)
   tail(cons(x0, x1))

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

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

(45)
YES