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


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


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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


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

(0) QTRS
(1) Overlay + Local Confluence [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QReductionProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QReductionProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (20) YES
    (21) QDP
        (22) UsableRulesProof [EQUIVALENT, 0 ms]
        (23) QDP
        (24) QReductionProof [EQUIVALENT, 0 ms]
        (25) QDP
        (26) QDPOrderProof [EQUIVALENT, 62 ms]
        (27) QDP
        (28) PisEmptyProof [EQUIVALENT, 0 ms]
        (29) YES
    (30) QDP
        (31) UsableRulesProof [EQUIVALENT, 0 ms]
        (32) QDP
        (33) QReductionProof [EQUIVALENT, 0 ms]
        (34) QDP
        (35) QDPOrderProof [EQUIVALENT, 0 ms]
        (36) QDP
        (37) PisEmptyProof [EQUIVALENT, 0 ms]
        (38) YES


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

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

   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   minus(0, y) -> 0
   minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
   if_minus(true, s(x), y) -> 0
   if_minus(false, s(x), y) -> s(minus(x, y))
   quot(0, s(y)) -> 0
   quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
   log(s(0)) -> 0
   log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

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:

   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   minus(0, y) -> 0
   minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
   if_minus(true, s(x), y) -> 0
   if_minus(false, s(x), y) -> s(minus(x, y))
   quot(0, s(y)) -> 0
   quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
   log(s(0)) -> 0
   log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

The set Q consists of the following terms:

   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   minus(0, x0)
   minus(s(x0), x1)
   if_minus(true, s(x0), x1)
   if_minus(false, s(x0), x1)
   quot(0, s(x0))
   quot(s(x0), s(x1))
   log(s(0))
   log(s(s(x0)))


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

(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:

   LE(s(x), s(y)) -> LE(x, y)
   MINUS(s(x), y) -> IF_MINUS(le(s(x), y), s(x), y)
   MINUS(s(x), y) -> LE(s(x), y)
   IF_MINUS(false, s(x), y) -> MINUS(x, y)
   QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))
   QUOT(s(x), s(y)) -> MINUS(x, y)
   LOG(s(s(x))) -> LOG(s(quot(x, s(s(0)))))
   LOG(s(s(x))) -> QUOT(x, s(s(0)))

The TRS R consists of the following rules:

   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   minus(0, y) -> 0
   minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
   if_minus(true, s(x), y) -> 0
   if_minus(false, s(x), y) -> s(minus(x, y))
   quot(0, s(y)) -> 0
   quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
   log(s(0)) -> 0
   log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

The set Q consists of the following terms:

   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   minus(0, x0)
   minus(s(x0), x1)
   if_minus(true, s(x0), x1)
   if_minus(false, s(x0), x1)
   quot(0, s(x0))
   quot(s(x0), s(x1))
   log(s(0))
   log(s(s(x0)))

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:

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

The TRS R consists of the following rules:

   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   minus(0, y) -> 0
   minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
   if_minus(true, s(x), y) -> 0
   if_minus(false, s(x), y) -> s(minus(x, y))
   quot(0, s(y)) -> 0
   quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
   log(s(0)) -> 0
   log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

The set Q consists of the following terms:

   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   minus(0, x0)
   minus(s(x0), x1)
   if_minus(true, s(x0), x1)
   if_minus(false, s(x0), x1)
   quot(0, s(x0))
   quot(s(x0), s(x1))
   log(s(0))
   log(s(s(x0)))

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

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

   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   minus(0, x0)
   minus(s(x0), x1)
   if_minus(true, s(x0), x1)
   if_minus(false, s(x0), x1)
   quot(0, s(x0))
   quot(s(x0), s(x1))
   log(s(0))
   log(s(s(x0)))

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].

   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   minus(0, x0)
   minus(s(x0), x1)
   if_minus(true, s(x0), x1)
   if_minus(false, s(x0), x1)
   quot(0, s(x0))
   quot(s(x0), s(x1))
   log(s(0))
   log(s(s(x0)))


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

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

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

   MINUS(s(x), y) -> IF_MINUS(le(s(x), y), s(x), y)
   IF_MINUS(false, s(x), y) -> MINUS(x, y)

The TRS R consists of the following rules:

   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   minus(0, y) -> 0
   minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
   if_minus(true, s(x), y) -> 0
   if_minus(false, s(x), y) -> s(minus(x, y))
   quot(0, s(y)) -> 0
   quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
   log(s(0)) -> 0
   log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

The set Q consists of the following terms:

   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   minus(0, x0)
   minus(s(x0), x1)
   if_minus(true, s(x0), x1)
   if_minus(false, s(x0), x1)
   quot(0, s(x0))
   quot(s(x0), s(x1))
   log(s(0))
   log(s(s(x0)))

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:

   MINUS(s(x), y) -> IF_MINUS(le(s(x), y), s(x), y)
   IF_MINUS(false, s(x), y) -> MINUS(x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   minus(0, x0)
   minus(s(x0), x1)
   if_minus(true, s(x0), x1)
   if_minus(false, s(x0), x1)
   quot(0, s(x0))
   quot(s(x0), s(x1))
   log(s(0))
   log(s(s(x0)))

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_minus(true, s(x0), x1)
   if_minus(false, s(x0), x1)
   quot(0, s(x0))
   quot(s(x0), s(x1))
   log(s(0))
   log(s(s(x0)))


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

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

   MINUS(s(x), y) -> IF_MINUS(le(s(x), y), s(x), y)
   IF_MINUS(false, s(x), y) -> MINUS(x, y)

The TRS R consists of the following rules:

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

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) 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_MINUS(false, s(x), y) -> MINUS(x, y)
The graph contains the following edges 2 > 1, 3 >= 2


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


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

(20)
YES

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

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

   QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))

The TRS R consists of the following rules:

   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   minus(0, y) -> 0
   minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
   if_minus(true, s(x), y) -> 0
   if_minus(false, s(x), y) -> s(minus(x, y))
   quot(0, s(y)) -> 0
   quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
   log(s(0)) -> 0
   log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

The set Q consists of the following terms:

   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   minus(0, x0)
   minus(s(x0), x1)
   if_minus(true, s(x0), x1)
   if_minus(false, s(x0), x1)
   quot(0, s(x0))
   quot(s(x0), s(x1))
   log(s(0))
   log(s(s(x0)))

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:

   QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))

The TRS R consists of the following rules:

   minus(0, y) -> 0
   minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   if_minus(true, s(x), y) -> 0
   if_minus(false, s(x), y) -> s(minus(x, y))
   le(0, y) -> true

The set Q consists of the following terms:

   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   minus(0, x0)
   minus(s(x0), x1)
   if_minus(true, s(x0), x1)
   if_minus(false, s(x0), x1)
   quot(0, s(x0))
   quot(s(x0), s(x1))
   log(s(0))
   log(s(s(x0)))

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].

   quot(0, s(x0))
   quot(s(x0), s(x1))
   log(s(0))
   log(s(s(x0)))


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

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

   QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))

The TRS R consists of the following rules:

   minus(0, y) -> 0
   minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   if_minus(true, s(x), y) -> 0
   if_minus(false, s(x), y) -> s(minus(x, y))
   le(0, y) -> true

The set Q consists of the following terms:

   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   minus(0, x0)
   minus(s(x0), x1)
   if_minus(true, s(x0), x1)
   if_minus(false, s(x0), x1)

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

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


The following pairs can be oriented strictly and are deleted.

   QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))
The remaining pairs can at least be oriented weakly.
Used ordering:  Combined order from the following AFS and order.
QUOT(x1, x2)  =  x1

s(x1)  =  s(x1)

minus(x1, x2)  =  x1

0  =  0

if_minus(x1, x2, x3)  =  x2


Knuth-Bendix order [KBO] with precedence:trivial

and weight map:

   s_1=1
   0=1

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

   minus(0, y) -> 0
   minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
   if_minus(false, s(x), y) -> s(minus(x, y))
   if_minus(true, s(x), y) -> 0


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

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

   minus(0, y) -> 0
   minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   if_minus(true, s(x), y) -> 0
   if_minus(false, s(x), y) -> s(minus(x, y))
   le(0, y) -> true

The set Q consists of the following terms:

   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   minus(0, x0)
   minus(s(x0), x1)
   if_minus(true, s(x0), x1)
   if_minus(false, s(x0), x1)

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

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

(29)
YES

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

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

   LOG(s(s(x))) -> LOG(s(quot(x, s(s(0)))))

The TRS R consists of the following rules:

   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   minus(0, y) -> 0
   minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
   if_minus(true, s(x), y) -> 0
   if_minus(false, s(x), y) -> s(minus(x, y))
   quot(0, s(y)) -> 0
   quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
   log(s(0)) -> 0
   log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

The set Q consists of the following terms:

   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   minus(0, x0)
   minus(s(x0), x1)
   if_minus(true, s(x0), x1)
   if_minus(false, s(x0), x1)
   quot(0, s(x0))
   quot(s(x0), s(x1))
   log(s(0))
   log(s(s(x0)))

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:

   LOG(s(s(x))) -> LOG(s(quot(x, s(s(0)))))

The TRS R consists of the following rules:

   quot(0, s(y)) -> 0
   quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
   minus(0, y) -> 0
   minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   if_minus(true, s(x), y) -> 0
   if_minus(false, s(x), y) -> s(minus(x, y))
   le(0, y) -> true

The set Q consists of the following terms:

   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   minus(0, x0)
   minus(s(x0), x1)
   if_minus(true, s(x0), x1)
   if_minus(false, s(x0), x1)
   quot(0, s(x0))
   quot(s(x0), s(x1))
   log(s(0))
   log(s(s(x0)))

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].

   log(s(0))
   log(s(s(x0)))


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

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

   LOG(s(s(x))) -> LOG(s(quot(x, s(s(0)))))

The TRS R consists of the following rules:

   quot(0, s(y)) -> 0
   quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
   minus(0, y) -> 0
   minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   if_minus(true, s(x), y) -> 0
   if_minus(false, s(x), y) -> s(minus(x, y))
   le(0, y) -> true

The set Q consists of the following terms:

   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   minus(0, x0)
   minus(s(x0), x1)
   if_minus(true, s(x0), x1)
   if_minus(false, s(x0), x1)
   quot(0, s(x0))
   quot(s(x0), s(x1))

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

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


The following pairs can be oriented strictly and are deleted.

   LOG(s(s(x))) -> LOG(s(quot(x, s(s(0)))))
The remaining pairs can at least be oriented weakly.
Used ordering:  Combined order from the following AFS and order.
LOG(x1)  =  x1

s(x1)  =  s(x1)

quot(x1, x2)  =  x1

0  =  0

minus(x1, x2)  =  x1

if_minus(x1, x2, x3)  =  x2


Knuth-Bendix order [KBO] with precedence:trivial

and weight map:

   s_1=1
   0=1

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

   quot(0, s(y)) -> 0
   quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
   minus(0, y) -> 0
   minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
   if_minus(false, s(x), y) -> s(minus(x, y))
   if_minus(true, s(x), y) -> 0


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

   quot(0, s(y)) -> 0
   quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
   minus(0, y) -> 0
   minus(s(x), y) -> if_minus(le(s(x), y), s(x), y)
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   if_minus(true, s(x), y) -> 0
   if_minus(false, s(x), y) -> s(minus(x, y))
   le(0, y) -> true

The set Q consists of the following terms:

   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   minus(0, x0)
   minus(s(x0), x1)
   if_minus(true, s(x0), x1)
   if_minus(false, s(x0), x1)
   quot(0, s(x0))
   quot(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
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(37) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
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(38)
YES