/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


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, 9 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) QDPOrderProof [EQUIVALENT, 0 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, 0 ms]
        (41) QDP
        (42) QDPOrderProof [EQUIVALENT, 85 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(0, nil)) -> 0
   min(cons(s(n), nil)) -> s(n)
   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))
   sort(nil) -> nil
   sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, x)))

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(0, nil)) -> 0
   min(cons(s(n), nil)) -> s(n)
   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))
   sort(nil) -> nil
   sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, 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(0, nil))
   min(cons(s(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))
   sort(nil)
   sort(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:

   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(cons(n, x)) -> MIN(cons(n, x))
   SORT(cons(n, x)) -> SORT(replace(min(cons(n, x)), n, x))
   SORT(cons(n, x)) -> REPLACE(min(cons(n, x)), n, 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(0, nil)) -> 0
   min(cons(s(n), nil)) -> s(n)
   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))
   sort(nil) -> nil
   sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, 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(0, nil))
   min(cons(s(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))
   sort(nil)
   sort(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 5 SCCs with 4 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(0, nil)) -> 0
   min(cons(s(n), nil)) -> s(n)
   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))
   sort(nil) -> nil
   sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, 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(0, nil))
   min(cons(s(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))
   sort(nil)
   sort(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:

   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(0, nil))
   min(cons(s(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))
   sort(nil)
   sort(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].

   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(0, nil))
   min(cons(s(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))
   sort(nil)
   sort(cons(x0, x1))


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

(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(0, nil)) -> 0
   min(cons(s(n), nil)) -> s(n)
   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))
   sort(nil) -> nil
   sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, 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(0, nil))
   min(cons(s(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))
   sort(nil)
   sort(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:

   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(0, nil))
   min(cons(s(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))
   sort(nil)
   sort(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].

   eq(0, 0)
   eq(0, s(x0))
   eq(s(x0), 0)
   eq(s(x0), s(x1))
   min(cons(0, nil))
   min(cons(s(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))
   sort(nil)
   sort(cons(x0, x1))


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

(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:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( IF_MIN_2(x_1, x_2) ) = 2x_2 + 2
POL( le_2(x_1, x_2) ) = 0
POL( 0 ) = 0
POL( true ) = 2
POL( s_1(x_1) ) = 2x_1 + 2
POL( false ) = 2
POL( MIN_1(x_1) ) = 2x_1 + 2
POL( cons_2(x_1, x_2) ) = 2x_2 + 2

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(0, nil)) -> 0
   min(cons(s(n), nil)) -> s(n)
   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))
   sort(nil) -> nil
   sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, 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(0, nil))
   min(cons(s(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))
   sort(nil)
   sort(cons(x0, x1))

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(0, nil))
   min(cons(s(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))
   sort(nil)
   sort(cons(x0, x1))

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(0, nil))
   min(cons(s(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))
   sort(nil)
   sort(cons(x0, x1))


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

(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(0, nil)) -> 0
   min(cons(s(n), nil)) -> s(n)
   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))
   sort(nil) -> nil
   sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, 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(0, nil))
   min(cons(s(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))
   sort(nil)
   sort(cons(x0, x1))

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(0, nil))
   min(cons(s(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))
   sort(nil)
   sort(cons(x0, x1))

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(0, nil))
   min(cons(s(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))
   sort(nil)
   sort(cons(x0, x1))


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

(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:

   SORT(cons(n, x)) -> SORT(replace(min(cons(n, x)), n, 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(0, nil)) -> 0
   min(cons(s(n), nil)) -> s(n)
   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))
   sort(nil) -> nil
   sort(cons(n, x)) -> cons(min(cons(n, x)), sort(replace(min(cons(n, x)), n, 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(0, nil))
   min(cons(s(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))
   sort(nil)
   sort(cons(x0, x1))

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:

   SORT(cons(n, x)) -> SORT(replace(min(cons(n, x)), n, x))

The TRS R consists of the following rules:

   min(cons(0, nil)) -> 0
   min(cons(s(n), nil)) -> s(n)
   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))
   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))
   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(0, nil))
   min(cons(s(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))
   sort(nil)
   sort(cons(x0, x1))

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(nil)
   sort(cons(x0, x1))


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

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

   SORT(cons(n, x)) -> SORT(replace(min(cons(n, x)), n, x))

The TRS R consists of the following rules:

   min(cons(0, nil)) -> 0
   min(cons(s(n), nil)) -> s(n)
   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))
   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))
   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(0, nil))
   min(cons(s(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))

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.

   SORT(cons(n, x)) -> SORT(replace(min(cons(n, x)), n, x))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( SORT_1(x_1) ) = 2x_1 + 2
POL( replace_3(x_1, ..., x_3) ) = 2x_2 + x_3 + 1
POL( min_1(x_1) ) = max{0, x_1 - 1}
POL( cons_2(x_1, x_2) ) = 2x_1 + x_2 + 2
POL( 0 ) = 2
POL( nil ) = 1
POL( s_1(x_1) ) = 2
POL( if_min_2(x_1, x_2) ) = max{0, 2x_1 + x_2 - 2}
POL( le_2(x_1, x_2) ) = x_1 + 2x_2 + 2
POL( true ) = 0
POL( false ) = 0
POL( if_replace_4(x_1, ..., x_4) ) = 2x_3 + x_4 + 1
POL( eq_2(x_1, x_2) ) = 0

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   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))


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

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

   min(cons(0, nil)) -> 0
   min(cons(s(n), nil)) -> s(n)
   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))
   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))
   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(0, nil))
   min(cons(s(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))

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