/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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


YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


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

(0) QTRS
(1) Overlay + Local Confluence [EQUIVALENT, 9 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, 1 ms]
        (27) QDP
        (28) QDPOrderProof [EQUIVALENT, 0 ms]
        (29) QDP
        (30) PisEmptyProof [EQUIVALENT, 0 ms]
        (31) YES


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

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

   minus(0, y) -> 0
   minus(s(x), 0) -> s(x)
   minus(s(x), s(y)) -> minus(x, y)
   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   if(true, x, y) -> x
   if(false, x, y) -> y
   perfectp(0) -> false
   perfectp(s(x)) -> f(x, s(0), s(x), s(x))
   f(0, y, 0, u) -> true
   f(0, y, s(z), u) -> false
   f(s(x), 0, z, u) -> f(x, u, minus(z, s(x)), u)
   f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

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), 0) -> s(x)
   minus(s(x), s(y)) -> minus(x, y)
   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   if(true, x, y) -> x
   if(false, x, y) -> y
   perfectp(0) -> false
   perfectp(s(x)) -> f(x, s(0), s(x), s(x))
   f(0, y, 0, u) -> true
   f(0, y, s(z), u) -> false
   f(s(x), 0, z, u) -> f(x, u, minus(z, s(x)), u)
   f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

   minus(0, x0)
   minus(s(x0), 0)
   minus(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   if(true, x0, x1)
   if(false, x0, x1)
   perfectp(0)
   perfectp(s(x0))
   f(0, x0, 0, x1)
   f(0, x0, s(x1), x2)
   f(s(x0), 0, x1, x2)
   f(s(x0), s(x1), x2, x3)


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

(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), s(y)) -> MINUS(x, y)
   LE(s(x), s(y)) -> LE(x, y)
   PERFECTP(s(x)) -> F(x, s(0), s(x), s(x))
   F(s(x), 0, z, u) -> F(x, u, minus(z, s(x)), u)
   F(s(x), 0, z, u) -> MINUS(z, s(x))
   F(s(x), s(y), z, u) -> IF(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))
   F(s(x), s(y), z, u) -> LE(x, y)
   F(s(x), s(y), z, u) -> F(s(x), minus(y, x), z, u)
   F(s(x), s(y), z, u) -> MINUS(y, x)
   F(s(x), s(y), z, u) -> F(x, u, z, u)

The TRS R consists of the following rules:

   minus(0, y) -> 0
   minus(s(x), 0) -> s(x)
   minus(s(x), s(y)) -> minus(x, y)
   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   if(true, x, y) -> x
   if(false, x, y) -> y
   perfectp(0) -> false
   perfectp(s(x)) -> f(x, s(0), s(x), s(x))
   f(0, y, 0, u) -> true
   f(0, y, s(z), u) -> false
   f(s(x), 0, z, u) -> f(x, u, minus(z, s(x)), u)
   f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

   minus(0, x0)
   minus(s(x0), 0)
   minus(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   if(true, x0, x1)
   if(false, x0, x1)
   perfectp(0)
   perfectp(s(x0))
   f(0, x0, 0, x1)
   f(0, x0, s(x1), x2)
   f(s(x0), 0, x1, x2)
   f(s(x0), s(x1), x2, x3)

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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 5 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:

   minus(0, y) -> 0
   minus(s(x), 0) -> s(x)
   minus(s(x), s(y)) -> minus(x, y)
   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   if(true, x, y) -> x
   if(false, x, y) -> y
   perfectp(0) -> false
   perfectp(s(x)) -> f(x, s(0), s(x), s(x))
   f(0, y, 0, u) -> true
   f(0, y, s(z), u) -> false
   f(s(x), 0, z, u) -> f(x, u, minus(z, s(x)), u)
   f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

   minus(0, x0)
   minus(s(x0), 0)
   minus(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   if(true, x0, x1)
   if(false, x0, x1)
   perfectp(0)
   perfectp(s(x0))
   f(0, x0, 0, x1)
   f(0, x0, s(x1), x2)
   f(s(x0), 0, x1, x2)
   f(s(x0), s(x1), x2, x3)

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:

   minus(0, x0)
   minus(s(x0), 0)
   minus(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   if(true, x0, x1)
   if(false, x0, x1)
   perfectp(0)
   perfectp(s(x0))
   f(0, x0, 0, x1)
   f(0, x0, s(x1), x2)
   f(s(x0), 0, x1, x2)
   f(s(x0), s(x1), x2, x3)

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), 0)
   minus(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   if(true, x0, x1)
   if(false, x0, x1)
   perfectp(0)
   perfectp(s(x0))
   f(0, x0, 0, x1)
   f(0, x0, s(x1), x2)
   f(s(x0), 0, x1, x2)
   f(s(x0), s(x1), x2, x3)


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

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

The TRS R consists of the following rules:

   minus(0, y) -> 0
   minus(s(x), 0) -> s(x)
   minus(s(x), s(y)) -> minus(x, y)
   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   if(true, x, y) -> x
   if(false, x, y) -> y
   perfectp(0) -> false
   perfectp(s(x)) -> f(x, s(0), s(x), s(x))
   f(0, y, 0, u) -> true
   f(0, y, s(z), u) -> false
   f(s(x), 0, z, u) -> f(x, u, minus(z, s(x)), u)
   f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

   minus(0, x0)
   minus(s(x0), 0)
   minus(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   if(true, x0, x1)
   if(false, x0, x1)
   perfectp(0)
   perfectp(s(x0))
   f(0, x0, 0, x1)
   f(0, x0, s(x1), x2)
   f(s(x0), 0, x1, x2)
   f(s(x0), s(x1), x2, x3)

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

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

   minus(0, x0)
   minus(s(x0), 0)
   minus(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   if(true, x0, x1)
   if(false, x0, x1)
   perfectp(0)
   perfectp(s(x0))
   f(0, x0, 0, x1)
   f(0, x0, s(x1), x2)
   f(s(x0), 0, x1, x2)
   f(s(x0), s(x1), x2, x3)

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), 0)
   minus(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   if(true, x0, x1)
   if(false, x0, x1)
   perfectp(0)
   perfectp(s(x0))
   f(0, x0, 0, x1)
   f(0, x0, s(x1), x2)
   f(s(x0), 0, x1, x2)
   f(s(x0), s(x1), x2, x3)


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

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

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

   F(s(x), s(y), z, u) -> F(s(x), minus(y, x), z, u)
   F(s(x), 0, z, u) -> F(x, u, minus(z, s(x)), u)
   F(s(x), s(y), z, u) -> F(x, u, z, u)

The TRS R consists of the following rules:

   minus(0, y) -> 0
   minus(s(x), 0) -> s(x)
   minus(s(x), s(y)) -> minus(x, y)
   le(0, y) -> true
   le(s(x), 0) -> false
   le(s(x), s(y)) -> le(x, y)
   if(true, x, y) -> x
   if(false, x, y) -> y
   perfectp(0) -> false
   perfectp(s(x)) -> f(x, s(0), s(x), s(x))
   f(0, y, 0, u) -> true
   f(0, y, s(z), u) -> false
   f(s(x), 0, z, u) -> f(x, u, minus(z, s(x)), u)
   f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

   minus(0, x0)
   minus(s(x0), 0)
   minus(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   if(true, x0, x1)
   if(false, x0, x1)
   perfectp(0)
   perfectp(s(x0))
   f(0, x0, 0, x1)
   f(0, x0, s(x1), x2)
   f(s(x0), 0, x1, x2)
   f(s(x0), s(x1), x2, x3)

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:

   F(s(x), s(y), z, u) -> F(s(x), minus(y, x), z, u)
   F(s(x), 0, z, u) -> F(x, u, minus(z, s(x)), u)
   F(s(x), s(y), z, u) -> F(x, u, z, u)

The TRS R consists of the following rules:

   minus(0, y) -> 0
   minus(s(x), s(y)) -> minus(x, y)
   minus(s(x), 0) -> s(x)

The set Q consists of the following terms:

   minus(0, x0)
   minus(s(x0), 0)
   minus(s(x0), s(x1))
   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   if(true, x0, x1)
   if(false, x0, x1)
   perfectp(0)
   perfectp(s(x0))
   f(0, x0, 0, x1)
   f(0, x0, s(x1), x2)
   f(s(x0), 0, x1, x2)
   f(s(x0), s(x1), x2, x3)

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].

   le(0, x0)
   le(s(x0), 0)
   le(s(x0), s(x1))
   if(true, x0, x1)
   if(false, x0, x1)
   perfectp(0)
   perfectp(s(x0))
   f(0, x0, 0, x1)
   f(0, x0, s(x1), x2)
   f(s(x0), 0, x1, x2)
   f(s(x0), s(x1), x2, x3)


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

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

   F(s(x), s(y), z, u) -> F(s(x), minus(y, x), z, u)
   F(s(x), 0, z, u) -> F(x, u, minus(z, s(x)), u)
   F(s(x), s(y), z, u) -> F(x, u, z, u)

The TRS R consists of the following rules:

   minus(0, y) -> 0
   minus(s(x), s(y)) -> minus(x, y)
   minus(s(x), 0) -> s(x)

The set Q consists of the following terms:

   minus(0, x0)
   minus(s(x0), 0)
   minus(s(x0), s(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.

   F(s(x), 0, z, u) -> F(x, u, minus(z, s(x)), u)
   F(s(x), s(y), z, u) -> F(x, u, z, u)
The remaining pairs can at least be oriented weakly.
Used ordering:  Combined order from the following AFS and order.
F(x1, x2, x3, x4)  =  x1

s(x1)  =  s(x1)


Knuth-Bendix order [KBO] with precedence:trivial

and weight map:

   s_1=1
   dummyConstant=1

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


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

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

   F(s(x), s(y), z, u) -> F(s(x), minus(y, x), z, u)

The TRS R consists of the following rules:

   minus(0, y) -> 0
   minus(s(x), s(y)) -> minus(x, y)
   minus(s(x), 0) -> s(x)

The set Q consists of the following terms:

   minus(0, x0)
   minus(s(x0), 0)
   minus(s(x0), s(x1))

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

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


The following pairs can be oriented strictly and are deleted.

   F(s(x), s(y), z, u) -> F(s(x), minus(y, x), z, u)
The remaining pairs can at least be oriented weakly.
Used ordering:  Combined order from the following AFS and order.
F(x1, x2, x3, x4)  =  x2

s(x1)  =  s(x1)

minus(x1, x2)  =  x1

0  =  0


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), s(y)) -> minus(x, y)
   minus(s(x), 0) -> s(x)


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

(29)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   minus(0, y) -> 0
   minus(s(x), s(y)) -> minus(x, y)
   minus(s(x), 0) -> s(x)

The set Q consists of the following terms:

   minus(0, x0)
   minus(s(x0), 0)
   minus(s(x0), s(x1))

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

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

(31)
YES