/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 13 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) TransformationProof [EQUIVALENT, 0 ms]
        (34) QDP
        (35) DependencyGraphProof [EQUIVALENT, 0 ms]
        (36) QDP
        (37) TransformationProof [EQUIVALENT, 0 ms]
        (38) QDP
        (39) DependencyGraphProof [EQUIVALENT, 0 ms]
        (40) QDP
        (41) TransformationProof [EQUIVALENT, 0 ms]
        (42) QDP
        (43) TransformationProof [EQUIVALENT, 0 ms]
        (44) QDP
        (45) Induction-Processor [SOUND, 45 ms]
        (46) AND
            (47) QDP
                (48) DependencyGraphProof [EQUIVALENT, 0 ms]
                (49) TRUE
            (50) QTRS
                (51) QTRSRRRProof [EQUIVALENT, 76 ms]
                (52) QTRS
                (53) RisEmptyProof [EQUIVALENT, 0 ms]
                (54) YES


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

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

   minus(0, y) -> 0
   minus(s(x), y) -> if(gt(s(x), y), x, y)
   if(true, x, y) -> s(minus(x, y))
   if(false, x, y) -> 0
   mod(x, 0) -> 0
   mod(x, s(y)) -> if1(lt(x, s(y)), x, s(y))
   if1(true, x, y) -> x
   if1(false, x, y) -> mod(minus(x, y), y)
   gt(0, y) -> false
   gt(s(x), 0) -> true
   gt(s(x), s(y)) -> gt(x, y)
   lt(x, 0) -> false
   lt(0, s(x)) -> true
   lt(s(x), s(y)) -> lt(x, y)

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:

   minus(0, y) -> 0
   minus(s(x), y) -> if(gt(s(x), y), x, y)
   if(true, x, y) -> s(minus(x, y))
   if(false, x, y) -> 0
   mod(x, 0) -> 0
   mod(x, s(y)) -> if1(lt(x, s(y)), x, s(y))
   if1(true, x, y) -> x
   if1(false, x, y) -> mod(minus(x, y), y)
   gt(0, y) -> false
   gt(s(x), 0) -> true
   gt(s(x), s(y)) -> gt(x, y)
   lt(x, 0) -> false
   lt(0, s(x)) -> true
   lt(s(x), s(y)) -> lt(x, y)

The set Q consists of the following terms:

   minus(0, x0)
   minus(s(x0), x1)
   if(true, x0, x1)
   if(false, x0, x1)
   mod(x0, 0)
   mod(x0, s(x1))
   if1(true, x0, x1)
   if1(false, x0, x1)
   gt(0, x0)
   gt(s(x0), 0)
   gt(s(x0), s(x1))
   lt(x0, 0)
   lt(0, s(x0))
   lt(s(x0), s(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:

   MINUS(s(x), y) -> IF(gt(s(x), y), x, y)
   MINUS(s(x), y) -> GT(s(x), y)
   IF(true, x, y) -> MINUS(x, y)
   MOD(x, s(y)) -> IF1(lt(x, s(y)), x, s(y))
   MOD(x, s(y)) -> LT(x, s(y))
   IF1(false, x, y) -> MOD(minus(x, y), y)
   IF1(false, x, y) -> MINUS(x, y)
   GT(s(x), s(y)) -> GT(x, y)
   LT(s(x), s(y)) -> LT(x, y)

The TRS R consists of the following rules:

   minus(0, y) -> 0
   minus(s(x), y) -> if(gt(s(x), y), x, y)
   if(true, x, y) -> s(minus(x, y))
   if(false, x, y) -> 0
   mod(x, 0) -> 0
   mod(x, s(y)) -> if1(lt(x, s(y)), x, s(y))
   if1(true, x, y) -> x
   if1(false, x, y) -> mod(minus(x, y), y)
   gt(0, y) -> false
   gt(s(x), 0) -> true
   gt(s(x), s(y)) -> gt(x, y)
   lt(x, 0) -> false
   lt(0, s(x)) -> true
   lt(s(x), s(y)) -> lt(x, y)

The set Q consists of the following terms:

   minus(0, x0)
   minus(s(x0), x1)
   if(true, x0, x1)
   if(false, x0, x1)
   mod(x0, 0)
   mod(x0, s(x1))
   if1(true, x0, x1)
   if1(false, x0, x1)
   gt(0, x0)
   gt(s(x0), 0)
   gt(s(x0), s(x1))
   lt(x0, 0)
   lt(0, s(x0))
   lt(s(x0), s(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:

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

The TRS R consists of the following rules:

   minus(0, y) -> 0
   minus(s(x), y) -> if(gt(s(x), y), x, y)
   if(true, x, y) -> s(minus(x, y))
   if(false, x, y) -> 0
   mod(x, 0) -> 0
   mod(x, s(y)) -> if1(lt(x, s(y)), x, s(y))
   if1(true, x, y) -> x
   if1(false, x, y) -> mod(minus(x, y), y)
   gt(0, y) -> false
   gt(s(x), 0) -> true
   gt(s(x), s(y)) -> gt(x, y)
   lt(x, 0) -> false
   lt(0, s(x)) -> true
   lt(s(x), s(y)) -> lt(x, y)

The set Q consists of the following terms:

   minus(0, x0)
   minus(s(x0), x1)
   if(true, x0, x1)
   if(false, x0, x1)
   mod(x0, 0)
   mod(x0, s(x1))
   if1(true, x0, x1)
   if1(false, x0, x1)
   gt(0, x0)
   gt(s(x0), 0)
   gt(s(x0), s(x1))
   lt(x0, 0)
   lt(0, s(x0))
   lt(s(x0), s(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:

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

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

   minus(0, x0)
   minus(s(x0), x1)
   if(true, x0, x1)
   if(false, x0, x1)
   mod(x0, 0)
   mod(x0, s(x1))
   if1(true, x0, x1)
   if1(false, x0, x1)
   gt(0, x0)
   gt(s(x0), 0)
   gt(s(x0), s(x1))
   lt(x0, 0)
   lt(0, s(x0))
   lt(s(x0), s(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].

   minus(0, x0)
   minus(s(x0), x1)
   if(true, x0, x1)
   if(false, x0, x1)
   mod(x0, 0)
   mod(x0, s(x1))
   if1(true, x0, x1)
   if1(false, x0, x1)
   gt(0, x0)
   gt(s(x0), 0)
   gt(s(x0), s(x1))
   lt(x0, 0)
   lt(0, s(x0))
   lt(s(x0), s(x1))


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

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

   LT(s(x), s(y)) -> LT(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:
*LT(s(x), s(y)) -> LT(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:

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

The TRS R consists of the following rules:

   minus(0, y) -> 0
   minus(s(x), y) -> if(gt(s(x), y), x, y)
   if(true, x, y) -> s(minus(x, y))
   if(false, x, y) -> 0
   mod(x, 0) -> 0
   mod(x, s(y)) -> if1(lt(x, s(y)), x, s(y))
   if1(true, x, y) -> x
   if1(false, x, y) -> mod(minus(x, y), y)
   gt(0, y) -> false
   gt(s(x), 0) -> true
   gt(s(x), s(y)) -> gt(x, y)
   lt(x, 0) -> false
   lt(0, s(x)) -> true
   lt(s(x), s(y)) -> lt(x, y)

The set Q consists of the following terms:

   minus(0, x0)
   minus(s(x0), x1)
   if(true, x0, x1)
   if(false, x0, x1)
   mod(x0, 0)
   mod(x0, s(x1))
   if1(true, x0, x1)
   if1(false, x0, x1)
   gt(0, x0)
   gt(s(x0), 0)
   gt(s(x0), s(x1))
   lt(x0, 0)
   lt(0, s(x0))
   lt(s(x0), s(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:

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

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

   minus(0, x0)
   minus(s(x0), x1)
   if(true, x0, x1)
   if(false, x0, x1)
   mod(x0, 0)
   mod(x0, s(x1))
   if1(true, x0, x1)
   if1(false, x0, x1)
   gt(0, x0)
   gt(s(x0), 0)
   gt(s(x0), s(x1))
   lt(x0, 0)
   lt(0, s(x0))
   lt(s(x0), s(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].

   minus(0, x0)
   minus(s(x0), x1)
   if(true, x0, x1)
   if(false, x0, x1)
   mod(x0, 0)
   mod(x0, s(x1))
   if1(true, x0, x1)
   if1(false, x0, x1)
   gt(0, x0)
   gt(s(x0), 0)
   gt(s(x0), s(x1))
   lt(x0, 0)
   lt(0, s(x0))
   lt(s(x0), s(x1))


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

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

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

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

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

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


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

(20)
YES

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

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

   IF(true, x, y) -> MINUS(x, y)
   MINUS(s(x), y) -> IF(gt(s(x), y), x, y)

The TRS R consists of the following rules:

   minus(0, y) -> 0
   minus(s(x), y) -> if(gt(s(x), y), x, y)
   if(true, x, y) -> s(minus(x, y))
   if(false, x, y) -> 0
   mod(x, 0) -> 0
   mod(x, s(y)) -> if1(lt(x, s(y)), x, s(y))
   if1(true, x, y) -> x
   if1(false, x, y) -> mod(minus(x, y), y)
   gt(0, y) -> false
   gt(s(x), 0) -> true
   gt(s(x), s(y)) -> gt(x, y)
   lt(x, 0) -> false
   lt(0, s(x)) -> true
   lt(s(x), s(y)) -> lt(x, y)

The set Q consists of the following terms:

   minus(0, x0)
   minus(s(x0), x1)
   if(true, x0, x1)
   if(false, x0, x1)
   mod(x0, 0)
   mod(x0, s(x1))
   if1(true, x0, x1)
   if1(false, x0, x1)
   gt(0, x0)
   gt(s(x0), 0)
   gt(s(x0), s(x1))
   lt(x0, 0)
   lt(0, s(x0))
   lt(s(x0), s(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:

   IF(true, x, y) -> MINUS(x, y)
   MINUS(s(x), y) -> IF(gt(s(x), y), x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   minus(0, x0)
   minus(s(x0), x1)
   if(true, x0, x1)
   if(false, x0, x1)
   mod(x0, 0)
   mod(x0, s(x1))
   if1(true, x0, x1)
   if1(false, x0, x1)
   gt(0, x0)
   gt(s(x0), 0)
   gt(s(x0), s(x1))
   lt(x0, 0)
   lt(0, s(x0))
   lt(s(x0), s(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].

   minus(0, x0)
   minus(s(x0), x1)
   if(true, x0, x1)
   if(false, x0, x1)
   mod(x0, 0)
   mod(x0, s(x1))
   if1(true, x0, x1)
   if1(false, x0, x1)
   lt(x0, 0)
   lt(0, s(x0))
   lt(s(x0), s(x1))


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

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

   IF(true, x, y) -> MINUS(x, y)
   MINUS(s(x), y) -> IF(gt(s(x), y), x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   gt(0, x0)
   gt(s(x0), 0)
   gt(s(x0), s(x1))

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:
*MINUS(s(x), y) -> IF(gt(s(x), y), x, y)
The graph contains the following edges 1 > 2, 2 >= 3


*IF(true, x, y) -> MINUS(x, y)
The graph contains the following edges 2 >= 1, 3 >= 2


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

(27)
YES

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

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

   IF1(false, x, y) -> MOD(minus(x, y), y)
   MOD(x, s(y)) -> IF1(lt(x, s(y)), x, s(y))

The TRS R consists of the following rules:

   minus(0, y) -> 0
   minus(s(x), y) -> if(gt(s(x), y), x, y)
   if(true, x, y) -> s(minus(x, y))
   if(false, x, y) -> 0
   mod(x, 0) -> 0
   mod(x, s(y)) -> if1(lt(x, s(y)), x, s(y))
   if1(true, x, y) -> x
   if1(false, x, y) -> mod(minus(x, y), y)
   gt(0, y) -> false
   gt(s(x), 0) -> true
   gt(s(x), s(y)) -> gt(x, y)
   lt(x, 0) -> false
   lt(0, s(x)) -> true
   lt(s(x), s(y)) -> lt(x, y)

The set Q consists of the following terms:

   minus(0, x0)
   minus(s(x0), x1)
   if(true, x0, x1)
   if(false, x0, x1)
   mod(x0, 0)
   mod(x0, s(x1))
   if1(true, x0, x1)
   if1(false, x0, x1)
   gt(0, x0)
   gt(s(x0), 0)
   gt(s(x0), s(x1))
   lt(x0, 0)
   lt(0, s(x0))
   lt(s(x0), s(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:

   IF1(false, x, y) -> MOD(minus(x, y), y)
   MOD(x, s(y)) -> IF1(lt(x, s(y)), x, s(y))

The TRS R consists of the following rules:

   lt(0, s(x)) -> true
   lt(s(x), s(y)) -> lt(x, y)
   lt(x, 0) -> false
   minus(0, y) -> 0
   minus(s(x), y) -> if(gt(s(x), y), x, y)
   gt(s(x), 0) -> true
   gt(s(x), s(y)) -> gt(x, y)
   if(true, x, y) -> s(minus(x, y))
   if(false, x, y) -> 0
   gt(0, y) -> false

The set Q consists of the following terms:

   minus(0, x0)
   minus(s(x0), x1)
   if(true, x0, x1)
   if(false, x0, x1)
   mod(x0, 0)
   mod(x0, s(x1))
   if1(true, x0, x1)
   if1(false, x0, x1)
   gt(0, x0)
   gt(s(x0), 0)
   gt(s(x0), s(x1))
   lt(x0, 0)
   lt(0, s(x0))
   lt(s(x0), s(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].

   mod(x0, 0)
   mod(x0, s(x1))
   if1(true, x0, x1)
   if1(false, x0, x1)


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

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

   IF1(false, x, y) -> MOD(minus(x, y), y)
   MOD(x, s(y)) -> IF1(lt(x, s(y)), x, s(y))

The TRS R consists of the following rules:

   lt(0, s(x)) -> true
   lt(s(x), s(y)) -> lt(x, y)
   lt(x, 0) -> false
   minus(0, y) -> 0
   minus(s(x), y) -> if(gt(s(x), y), x, y)
   gt(s(x), 0) -> true
   gt(s(x), s(y)) -> gt(x, y)
   if(true, x, y) -> s(minus(x, y))
   if(false, x, y) -> 0
   gt(0, y) -> false

The set Q consists of the following terms:

   minus(0, x0)
   minus(s(x0), x1)
   if(true, x0, x1)
   if(false, x0, x1)
   gt(0, x0)
   gt(s(x0), 0)
   gt(s(x0), s(x1))
   lt(x0, 0)
   lt(0, s(x0))
   lt(s(x0), s(x1))

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

(33) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule MOD(x, s(y)) -> IF1(lt(x, s(y)), x, s(y)) at position [0] we obtained the following new rules [LPAR04]:

   (MOD(0, s(x0)) -> IF1(true, 0, s(x0)),MOD(0, s(x0)) -> IF1(true, 0, s(x0)))
   (MOD(s(x0), s(x1)) -> IF1(lt(x0, x1), s(x0), s(x1)),MOD(s(x0), s(x1)) -> IF1(lt(x0, x1), s(x0), s(x1)))


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

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

   IF1(false, x, y) -> MOD(minus(x, y), y)
   MOD(0, s(x0)) -> IF1(true, 0, s(x0))
   MOD(s(x0), s(x1)) -> IF1(lt(x0, x1), s(x0), s(x1))

The TRS R consists of the following rules:

   lt(0, s(x)) -> true
   lt(s(x), s(y)) -> lt(x, y)
   lt(x, 0) -> false
   minus(0, y) -> 0
   minus(s(x), y) -> if(gt(s(x), y), x, y)
   gt(s(x), 0) -> true
   gt(s(x), s(y)) -> gt(x, y)
   if(true, x, y) -> s(minus(x, y))
   if(false, x, y) -> 0
   gt(0, y) -> false

The set Q consists of the following terms:

   minus(0, x0)
   minus(s(x0), x1)
   if(true, x0, x1)
   if(false, x0, x1)
   gt(0, x0)
   gt(s(x0), 0)
   gt(s(x0), s(x1))
   lt(x0, 0)
   lt(0, s(x0))
   lt(s(x0), s(x1))

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

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

(36)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MOD(s(x0), s(x1)) -> IF1(lt(x0, x1), s(x0), s(x1))
   IF1(false, x, y) -> MOD(minus(x, y), y)

The TRS R consists of the following rules:

   lt(0, s(x)) -> true
   lt(s(x), s(y)) -> lt(x, y)
   lt(x, 0) -> false
   minus(0, y) -> 0
   minus(s(x), y) -> if(gt(s(x), y), x, y)
   gt(s(x), 0) -> true
   gt(s(x), s(y)) -> gt(x, y)
   if(true, x, y) -> s(minus(x, y))
   if(false, x, y) -> 0
   gt(0, y) -> false

The set Q consists of the following terms:

   minus(0, x0)
   minus(s(x0), x1)
   if(true, x0, x1)
   if(false, x0, x1)
   gt(0, x0)
   gt(s(x0), 0)
   gt(s(x0), s(x1))
   lt(x0, 0)
   lt(0, s(x0))
   lt(s(x0), s(x1))

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

(37) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule IF1(false, x, y) -> MOD(minus(x, y), y) at position [0] we obtained the following new rules [LPAR04]:

   (IF1(false, 0, x0) -> MOD(0, x0),IF1(false, 0, x0) -> MOD(0, x0))
   (IF1(false, s(x0), x1) -> MOD(if(gt(s(x0), x1), x0, x1), x1),IF1(false, s(x0), x1) -> MOD(if(gt(s(x0), x1), x0, x1), x1))


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

(38)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MOD(s(x0), s(x1)) -> IF1(lt(x0, x1), s(x0), s(x1))
   IF1(false, 0, x0) -> MOD(0, x0)
   IF1(false, s(x0), x1) -> MOD(if(gt(s(x0), x1), x0, x1), x1)

The TRS R consists of the following rules:

   lt(0, s(x)) -> true
   lt(s(x), s(y)) -> lt(x, y)
   lt(x, 0) -> false
   minus(0, y) -> 0
   minus(s(x), y) -> if(gt(s(x), y), x, y)
   gt(s(x), 0) -> true
   gt(s(x), s(y)) -> gt(x, y)
   if(true, x, y) -> s(minus(x, y))
   if(false, x, y) -> 0
   gt(0, y) -> false

The set Q consists of the following terms:

   minus(0, x0)
   minus(s(x0), x1)
   if(true, x0, x1)
   if(false, x0, x1)
   gt(0, x0)
   gt(s(x0), 0)
   gt(s(x0), s(x1))
   lt(x0, 0)
   lt(0, s(x0))
   lt(s(x0), s(x1))

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

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

(40)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF1(false, s(x0), x1) -> MOD(if(gt(s(x0), x1), x0, x1), x1)
   MOD(s(x0), s(x1)) -> IF1(lt(x0, x1), s(x0), s(x1))

The TRS R consists of the following rules:

   lt(0, s(x)) -> true
   lt(s(x), s(y)) -> lt(x, y)
   lt(x, 0) -> false
   minus(0, y) -> 0
   minus(s(x), y) -> if(gt(s(x), y), x, y)
   gt(s(x), 0) -> true
   gt(s(x), s(y)) -> gt(x, y)
   if(true, x, y) -> s(minus(x, y))
   if(false, x, y) -> 0
   gt(0, y) -> false

The set Q consists of the following terms:

   minus(0, x0)
   minus(s(x0), x1)
   if(true, x0, x1)
   if(false, x0, x1)
   gt(0, x0)
   gt(s(x0), 0)
   gt(s(x0), s(x1))
   lt(x0, 0)
   lt(0, s(x0))
   lt(s(x0), s(x1))

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

(41) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule IF1(false, s(x0), x1) -> MOD(if(gt(s(x0), x1), x0, x1), x1) we obtained the following new rules [LPAR04]:

   (IF1(false, s(z0), s(z1)) -> MOD(if(gt(s(z0), s(z1)), z0, s(z1)), s(z1)),IF1(false, s(z0), s(z1)) -> MOD(if(gt(s(z0), s(z1)), z0, s(z1)), s(z1)))


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

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

   MOD(s(x0), s(x1)) -> IF1(lt(x0, x1), s(x0), s(x1))
   IF1(false, s(z0), s(z1)) -> MOD(if(gt(s(z0), s(z1)), z0, s(z1)), s(z1))

The TRS R consists of the following rules:

   lt(0, s(x)) -> true
   lt(s(x), s(y)) -> lt(x, y)
   lt(x, 0) -> false
   minus(0, y) -> 0
   minus(s(x), y) -> if(gt(s(x), y), x, y)
   gt(s(x), 0) -> true
   gt(s(x), s(y)) -> gt(x, y)
   if(true, x, y) -> s(minus(x, y))
   if(false, x, y) -> 0
   gt(0, y) -> false

The set Q consists of the following terms:

   minus(0, x0)
   minus(s(x0), x1)
   if(true, x0, x1)
   if(false, x0, x1)
   gt(0, x0)
   gt(s(x0), 0)
   gt(s(x0), s(x1))
   lt(x0, 0)
   lt(0, s(x0))
   lt(s(x0), s(x1))

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

(43) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule IF1(false, s(z0), s(z1)) -> MOD(if(gt(s(z0), s(z1)), z0, s(z1)), s(z1)) at position [0,0] we obtained the following new rules [LPAR04]:

   (IF1(false, s(z0), s(z1)) -> MOD(if(gt(z0, z1), z0, s(z1)), s(z1)),IF1(false, s(z0), s(z1)) -> MOD(if(gt(z0, z1), z0, s(z1)), s(z1)))


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

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

   MOD(s(x0), s(x1)) -> IF1(lt(x0, x1), s(x0), s(x1))
   IF1(false, s(z0), s(z1)) -> MOD(if(gt(z0, z1), z0, s(z1)), s(z1))

The TRS R consists of the following rules:

   lt(0, s(x)) -> true
   lt(s(x), s(y)) -> lt(x, y)
   lt(x, 0) -> false
   minus(0, y) -> 0
   minus(s(x), y) -> if(gt(s(x), y), x, y)
   gt(s(x), 0) -> true
   gt(s(x), s(y)) -> gt(x, y)
   if(true, x, y) -> s(minus(x, y))
   if(false, x, y) -> 0
   gt(0, y) -> false

The set Q consists of the following terms:

   minus(0, x0)
   minus(s(x0), x1)
   if(true, x0, x1)
   if(false, x0, x1)
   gt(0, x0)
   gt(s(x0), 0)
   gt(s(x0), s(x1))
   lt(x0, 0)
   lt(0, s(x0))
   lt(s(x0), s(x1))

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

(45) Induction-Processor (SOUND)

This DP could be deleted by the Induction-Processor:
IF1(false_renamed, s(z0), s(z1)) -> MOD(if(gt(z0, z1), z0, s(z1)), s(z1))


This order was computed:
Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(IF1(x_1, x_2, x_3)) = x_2
   POL(MOD(x_1, x_2)) = x_1
   POL(false_renamed) = 0
   POL(gt(x_1, x_2)) = 0
   POL(if(x_1, x_2, x_3)) = 1 + x_2
   POL(lt(x_1, x_2)) = x_1
   POL(minus(x_1, x_2)) = x_1
   POL(s(x_1)) = 1 + x_1
   POL(true_renamed) = 0

At least one of these decreasing rules is always used after the deleted DP:
if(false_renamed, x5, y1) -> 0


The following formula is valid:
z0:sort[a22],z1:sort[a22].if'(gt(z0, z1), z0, s(z1))=true


The transformed set:
if'(true_renamed, x4, y'') -> minus'(x4, y'')
if'(false_renamed, x5, y1) -> true
minus'(0, y3) -> false
minus'(s(x6), y4) -> if'(gt(s(x6), y4), x6, y4)
lt(0, s(x)) -> true_renamed
lt(s(x'), s(y)) -> lt(x', y)
lt(x'', 0) -> false_renamed
gt(s(x2), 0) -> true_renamed
gt(s(x3), s(y')) -> gt(x3, y')
if(true_renamed, x4, y'') -> s(minus(x4, y''))
if(false_renamed, x5, y1) -> 0
gt(0, y2) -> false_renamed
minus(0, y3) -> 0
minus(s(x6), y4) -> if(gt(s(x6), y4), x6, y4)
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[a22](0, 0) -> true
equal_sort[a22](0, s(v30)) -> false
equal_sort[a22](s(v31), 0) -> false
equal_sort[a22](s(v31), s(v32)) -> equal_sort[a22](v31, v32)
equal_sort[a21](true_renamed, true_renamed) -> true
equal_sort[a21](true_renamed, false_renamed) -> false
equal_sort[a21](false_renamed, true_renamed) -> false
equal_sort[a21](false_renamed, false_renamed) -> true
equal_sort[a39](witness_sort[a39], witness_sort[a39]) -> 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, v30, v31, v32, x5, y1, y'', x6, y3, y4, x', y, x'', x2, x3, y', y2]
   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[a22](0, 0) -> true
   equal_sort[a22](0, s(v30)) -> false
   equal_sort[a22](s(v31), 0) -> false
   equal_sort[a22](s(v31), s(v32)) -> equal_sort[a22](v31, v32)
   equal_sort[a21](true_renamed, true_renamed) -> true
   equal_sort[a21](true_renamed, false_renamed) -> false
   equal_sort[a21](false_renamed, true_renamed) -> false
   equal_sort[a21](false_renamed, false_renamed) -> true
   equal_sort[a39](witness_sort[a39], witness_sort[a39]) -> true
   if'(false_renamed, x5, y1) -> true
   if'(true_renamed, 0, y'') -> false
   if'(true_renamed, s(x6), y'') -> if'(gt(s(x6), y''), x6, y'')
   minus'(0, y3) -> false
   equal_sort[a21](gt(s(x6), y4), true_renamed) -> true | minus'(s(x6), y4) -> minus'(x6, y4)
   equal_sort[a21](gt(s(x6), y4), true_renamed) -> false | minus'(s(x6), y4) -> true
   lt(0, s(x)) -> true_renamed
   lt(s(x'), s(y)) -> lt(x', y)
   lt(x'', 0) -> false_renamed
   gt(s(x2), 0) -> true_renamed
   gt(s(x3), s(y')) -> gt(x3, y')
   gt(0, y2) -> false_renamed
   if(false_renamed, x5, y1) -> 0
   if(true_renamed, 0, y'') -> s(0)
   if(true_renamed, s(x6), y'') -> s(if(gt(s(x6), y''), x6, y''))
   minus(0, y3) -> 0
   equal_sort[a21](gt(s(x6), y4), true_renamed) -> true | minus(s(x6), y4) -> s(minus(x6, y4))
   equal_sort[a21](gt(s(x6), y4), true_renamed) -> false | minus(s(x6), y4) -> 0

using the following formula:
z0:sort[a22],z1:sort[a22].if'(gt(z0, z1), z0, s(z1))=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) Symbolic evaluation under hypothesis [EQUIVALENT, 0 ms]
                (13) YES
    (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) Inverse Substitution [SOUND, 0 ms]
        (23) Formula
        (24) Inverse Substitution [SOUND, 0 ms]
        (25) Formula
        (26) Induction by algorithm [EQUIVALENT, 0 ms]
        (27) AND
            (28) Formula
                (29) Symbolic evaluation [EQUIVALENT, 0 ms]
                (30) YES
            (31) Formula
                (32) Symbolic evaluation [EQUIVALENT, 0 ms]
                (33) YES
            (34) Formula
                (35) Symbolic evaluation under hypothesis [EQUIVALENT, 0 ms]
                (36) YES


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

(0)
Obligation:
Formula:
z0:sort[a22],z1:sort[a22].if'(gt(z0, z1), z0, s(z1))=true

There are no hypotheses.




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

(1) Induction by algorithm (EQUIVALENT)
Induction by algorithm gt(z0, z1) generates the following cases:

1. Base Case:
Formula:
x2:sort[a22].if'(gt(s(x2), 0), s(x2), s(0))=true

There are no hypotheses.





2. Base Case:
Formula:
y2:sort[a22].if'(gt(0, y2), 0, s(y2))=true

There are no hypotheses.





1. Step Case:
Formula:
x3:sort[a22],y':sort[a22].if'(gt(s(x3), s(y')), s(x3), s(s(y')))=true

Hypotheses:
x3:sort[a22],y':sort[a22].if'(gt(x3, y'), x3, s(y'))=true






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

(2)
Complex Obligation (AND)

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

(3)
Obligation:
Formula:
x2:sort[a22].if'(gt(s(x2), 0), s(x2), s(0))=true

There are no hypotheses.




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

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

(5)
Obligation:
Formula:
x2:sort[a22].if'(gt(x2, 0), x2, s(0))=true

There are no hypotheses.




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

(6) Induction by data structure (EQUIVALENT)
Induction by data structure sort[a22] generates the following cases:



1. Base Case:
Formula:
if'(gt(0, 0), 0, s(0))=true

There are no hypotheses.





1. Step Case:
Formula:
n:sort[a22].if'(gt(s(n), 0), s(n), s(0))=true

Hypotheses:
n:sort[a22].if'(gt(n, 0), n, s(0))=true






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

(7)
Complex Obligation (AND)

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

(8)
Obligation:
Formula:
if'(gt(0, 0), 0, s(0))=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[a22].if'(gt(s(n), 0), s(n), s(0))=true

Hypotheses:
n:sort[a22].if'(gt(n, 0), n, s(0))=true




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

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

n:sort[a22].if'(gt(n, 0), n, s(0))=true

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

(13)
YES

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

(14)
Obligation:
Formula:
y2:sort[a22].if'(gt(0, y2), 0, s(y2))=true

There are no hypotheses.




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

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

(16)
YES

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

(17)
Obligation:
Formula:
x3:sort[a22],y':sort[a22].if'(gt(s(x3), s(y')), s(x3), s(s(y')))=true

Hypotheses:
x3:sort[a22],y':sort[a22].if'(gt(x3, y'), x3, s(y'))=true




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

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

(19)
Obligation:
Formula:
x3:sort[a22],y':sort[a22].if'(gt(x3, y'), s(x3), s(s(y')))=true

Hypotheses:
x3:sort[a22],y':sort[a22].if'(gt(x3, y'), x3, s(y'))=true




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

(20) Hypothesis Lifting (EQUIVALENT)
Formula could be generalised by hypothesis lifting to the following new obligation:
Formula:
x3:sort[a22],y':sort[a22].(if'(gt(x3, y'), x3, s(y'))=true->if'(gt(x3, y'), s(x3), s(s(y')))=true)

There are no hypotheses.




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

(21)
Obligation:
Formula:
x3:sort[a22],y':sort[a22].(if'(gt(x3, y'), x3, s(y'))=true->if'(gt(x3, y'), s(x3), s(s(y')))=true)

There are no hypotheses.




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

(22) Inverse Substitution (SOUND)
The formula could be generalised by inverse substitution to:
n:sort[a21],x3:sort[a22],y':sort[a22].(if'(n, x3, s(y'))=true->if'(n, s(x3), s(s(y')))=true)

Inverse substitution used:
[gt(x3, y')/n]


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

(23)
Obligation:
Formula:
n:sort[a21],x3:sort[a22],y':sort[a22].(if'(n, x3, s(y'))=true->if'(n, s(x3), s(s(y')))=true)

There are no hypotheses.




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

(24) Inverse Substitution (SOUND)
The formula could be generalised by inverse substitution to:
n:sort[a21],x3:sort[a22],n':sort[a22].(if'(n, x3, n')=true->if'(n, s(x3), s(n'))=true)

Inverse substitution used:
[s(y')/n']


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

(25)
Obligation:
Formula:
n:sort[a21],x3:sort[a22],n':sort[a22].(if'(n, x3, n')=true->if'(n, s(x3), s(n'))=true)

There are no hypotheses.




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

(26) Induction by algorithm (EQUIVALENT)
Induction by algorithm if'(n, x3, n') generates the following cases:

1. Base Case:
Formula:
x5:sort[a22],y1:sort[a22].(if'(false_renamed, x5, y1)=true->if'(false_renamed, s(x5), s(y1))=true)

There are no hypotheses.





2. Base Case:
Formula:
y'':sort[a22].(if'(true_renamed, 0, y'')=true->if'(true_renamed, s(0), s(y''))=true)

There are no hypotheses.





1. Step Case:
Formula:
x6:sort[a22],y'':sort[a22].(if'(true_renamed, s(x6), y'')=true->if'(true_renamed, s(s(x6)), s(y''))=true)

Hypotheses:
x6:sort[a22],y'':sort[a22].(if'(gt(s(x6), y''), x6, y'')=true->if'(gt(s(x6), y''), s(x6), s(y''))=true)






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

(27)
Complex Obligation (AND)

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

(28)
Obligation:
Formula:
x5:sort[a22],y1:sort[a22].(if'(false_renamed, x5, y1)=true->if'(false_renamed, s(x5), s(y1))=true)

There are no hypotheses.




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

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

(30)
YES

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

(31)
Obligation:
Formula:
y'':sort[a22].(if'(true_renamed, 0, y'')=true->if'(true_renamed, s(0), s(y''))=true)

There are no hypotheses.




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

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

(33)
YES

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

(34)
Obligation:
Formula:
x6:sort[a22],y'':sort[a22].(if'(true_renamed, s(x6), y'')=true->if'(true_renamed, s(s(x6)), s(y''))=true)

Hypotheses:
x6:sort[a22],y'':sort[a22].(if'(gt(s(x6), y''), x6, y'')=true->if'(gt(s(x6), y''), s(x6), s(y''))=true)




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

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

x6:sort[a22],y'':sort[a22].(if'(gt(s(x6), y''), x6, y'')=true->if'(gt(s(x6), y''), s(x6), s(y''))=true)

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

(36)
YES

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

(46)
Complex Obligation (AND)

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

(47)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MOD(s(x0), s(x1)) -> IF1(lt(x0, x1), s(x0), s(x1))

The TRS R consists of the following rules:

   lt(0, s(x)) -> true
   lt(s(x), s(y)) -> lt(x, y)
   lt(x, 0) -> false
   minus(0, y) -> 0
   minus(s(x), y) -> if(gt(s(x), y), x, y)
   gt(s(x), 0) -> true
   gt(s(x), s(y)) -> gt(x, y)
   if(true, x, y) -> s(minus(x, y))
   if(false, x, y) -> 0
   gt(0, y) -> false

The set Q consists of the following terms:

   minus(0, x0)
   minus(s(x0), x1)
   if(true, x0, x1)
   if(false, x0, x1)
   gt(0, x0)
   gt(s(x0), 0)
   gt(s(x0), s(x1))
   lt(x0, 0)
   lt(0, s(x0))
   lt(s(x0), s(x1))

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

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

(49)
TRUE

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

(50)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   if'(true_renamed, x4, y'') -> minus'(x4, y'')
   if'(false_renamed, x5, y1) -> true
   minus'(0, y3) -> false
   minus'(s(x6), y4) -> if'(gt(s(x6), y4), x6, y4)
   lt(0, s(x)) -> true_renamed
   lt(s(x'), s(y)) -> lt(x', y)
   lt(x'', 0) -> false_renamed
   gt(s(x2), 0) -> true_renamed
   gt(s(x3), s(y')) -> gt(x3, y')
   if(true_renamed, x4, y'') -> s(minus(x4, y''))
   if(false_renamed, x5, y1) -> 0
   gt(0, y2) -> false_renamed
   minus(0, y3) -> 0
   minus(s(x6), y4) -> if(gt(s(x6), y4), x6, y4)
   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[a22](0, 0) -> true
   equal_sort[a22](0, s(v30)) -> false
   equal_sort[a22](s(v31), 0) -> false
   equal_sort[a22](s(v31), s(v32)) -> equal_sort[a22](v31, v32)
   equal_sort[a21](true_renamed, true_renamed) -> true
   equal_sort[a21](true_renamed, false_renamed) -> false
   equal_sort[a21](false_renamed, true_renamed) -> false
   equal_sort[a21](false_renamed, false_renamed) -> true
   equal_sort[a39](witness_sort[a39], witness_sort[a39]) -> true

Q is empty.

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

(51) QTRSRRRProof (EQUIVALENT)
Used ordering:
Quasi precedence:
[if'_3, minus'_2] > [true_renamed, gt_2] > true > false
[if'_3, minus'_2] > [true_renamed, gt_2] > s_1 > false
[if'_3, minus'_2] > [true_renamed, gt_2] > s_1 > lt_2
[false_renamed, 0, if_3, minus_2] > [true_renamed, gt_2] > true > false
[false_renamed, 0, if_3, minus_2] > [true_renamed, gt_2] > s_1 > false
[false_renamed, 0, if_3, minus_2] > [true_renamed, gt_2] > s_1 > lt_2
equal_bool_2 > true > false
not_1 > true > false
isa_false_1 > true > false
equal_sort[a22]_2 > true > false
equal_sort[a21]_2 > true > false
equal_sort[a39]_2 > true > false
witness_sort[a39] > true > false


Status:
if'_3: [2,3,1]
true_renamed: multiset status
minus'_2: [1,2]
false_renamed: multiset status
true: multiset status
0: multiset status
false: multiset status
s_1: [1]
gt_2: multiset status
lt_2: [1,2]
if_3: [2,3,1]
minus_2: [1,2]
equal_bool_2: multiset status
and_2: [2,1]
or_2: [2,1]
not_1: multiset status
isa_true_1: multiset status
isa_false_1: multiset status
equal_sort[a22]_2: multiset status
equal_sort[a21]_2: [1,2]
equal_sort[a39]_2: multiset status
witness_sort[a39]: multiset status

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

   if'(true_renamed, x4, y'') -> minus'(x4, y'')
   if'(false_renamed, x5, y1) -> true
   minus'(0, y3) -> false
   minus'(s(x6), y4) -> if'(gt(s(x6), y4), x6, y4)
   lt(0, s(x)) -> true_renamed
   lt(s(x'), s(y)) -> lt(x', y)
   lt(x'', 0) -> false_renamed
   gt(s(x2), 0) -> true_renamed
   gt(s(x3), s(y')) -> gt(x3, y')
   if(true_renamed, x4, y'') -> s(minus(x4, y''))
   if(false_renamed, x5, y1) -> 0
   gt(0, y2) -> false_renamed
   minus(0, y3) -> 0
   minus(s(x6), y4) -> if(gt(s(x6), y4), x6, y4)
   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[a22](0, 0) -> true
   equal_sort[a22](0, s(v30)) -> false
   equal_sort[a22](s(v31), 0) -> false
   equal_sort[a22](s(v31), s(v32)) -> equal_sort[a22](v31, v32)
   equal_sort[a21](true_renamed, true_renamed) -> true
   equal_sort[a21](true_renamed, false_renamed) -> false
   equal_sort[a21](false_renamed, true_renamed) -> false
   equal_sort[a21](false_renamed, false_renamed) -> true
   equal_sort[a39](witness_sort[a39], witness_sort[a39]) -> true




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

(52)
Obligation:
Q restricted rewrite system:
R is empty.
Q is empty.

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

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

(54)
YES