/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (9) YES
    (10) QDP
        (11) UsableRulesProof [EQUIVALENT, 0 ms]
        (12) QDP
        (13) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (14) YES
    (15) QDP
        (16) UsableRulesProof [EQUIVALENT, 0 ms]
        (17) QDP
        (18) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (19) YES
    (20) QDP
        (21) QDPOrderProof [EQUIVALENT, 31 ms]
        (22) QDP
        (23) QDPOrderProof [EQUIVALENT, 0 ms]
        (24) QDP
        (25) TransformationProof [EQUIVALENT, 0 ms]
        (26) QDP
        (27) DependencyGraphProof [EQUIVALENT, 0 ms]
        (28) TRUE
    (29) QDP
        (30) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (31) YES


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

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

   plus(x, 0) -> x
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   times(0, y) -> 0
   times(s(0), y) -> y
   times(s(x), y) -> plus(y, times(x, y))
   div(0, y) -> 0
   div(x, y) -> quot(x, y, y)
   quot(0, s(y), z) -> 0
   quot(s(x), s(y), z) -> quot(x, y, z)
   quot(x, 0, s(z)) -> s(div(x, s(z)))
   div(div(x, y), z) -> div(x, times(y, z))
   eq(0, 0) -> true
   eq(s(x), 0) -> false
   eq(0, s(y)) -> false
   eq(s(x), s(y)) -> eq(x, y)
   divides(y, x) -> eq(x, times(div(x, y), y))
   prime(s(s(x))) -> pr(s(s(x)), s(x))
   pr(x, s(0)) -> true
   pr(x, s(s(y))) -> if(divides(s(s(y)), x), x, s(y))
   if(true, x, y) -> false
   if(false, x, y) -> pr(x, y)

Q is empty.

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

(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   PLUS(s(x), y) -> PLUS(x, y)
   TIMES(s(x), y) -> PLUS(y, times(x, y))
   TIMES(s(x), y) -> TIMES(x, y)
   DIV(x, y) -> QUOT(x, y, y)
   QUOT(s(x), s(y), z) -> QUOT(x, y, z)
   QUOT(x, 0, s(z)) -> DIV(x, s(z))
   DIV(div(x, y), z) -> DIV(x, times(y, z))
   DIV(div(x, y), z) -> TIMES(y, z)
   EQ(s(x), s(y)) -> EQ(x, y)
   DIVIDES(y, x) -> EQ(x, times(div(x, y), y))
   DIVIDES(y, x) -> TIMES(div(x, y), y)
   DIVIDES(y, x) -> DIV(x, y)
   PRIME(s(s(x))) -> PR(s(s(x)), s(x))
   PR(x, s(s(y))) -> IF(divides(s(s(y)), x), x, s(y))
   PR(x, s(s(y))) -> DIVIDES(s(s(y)), x)
   IF(false, x, y) -> PR(x, y)

The TRS R consists of the following rules:

   plus(x, 0) -> x
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   times(0, y) -> 0
   times(s(0), y) -> y
   times(s(x), y) -> plus(y, times(x, y))
   div(0, y) -> 0
   div(x, y) -> quot(x, y, y)
   quot(0, s(y), z) -> 0
   quot(s(x), s(y), z) -> quot(x, y, z)
   quot(x, 0, s(z)) -> s(div(x, s(z)))
   div(div(x, y), z) -> div(x, times(y, z))
   eq(0, 0) -> true
   eq(s(x), 0) -> false
   eq(0, s(y)) -> false
   eq(s(x), s(y)) -> eq(x, y)
   divides(y, x) -> eq(x, times(div(x, y), y))
   prime(s(s(x))) -> pr(s(s(x)), s(x))
   pr(x, s(0)) -> true
   pr(x, s(s(y))) -> if(divides(s(s(y)), x), x, s(y))
   if(true, x, y) -> false
   if(false, x, y) -> pr(x, y)

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

(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 5 SCCs with 7 less nodes.
----------------------------------------

(4)
Complex Obligation (AND)

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

(5)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

   plus(x, 0) -> x
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   times(0, y) -> 0
   times(s(0), y) -> y
   times(s(x), y) -> plus(y, times(x, y))
   div(0, y) -> 0
   div(x, y) -> quot(x, y, y)
   quot(0, s(y), z) -> 0
   quot(s(x), s(y), z) -> quot(x, y, z)
   quot(x, 0, s(z)) -> s(div(x, s(z)))
   div(div(x, y), z) -> div(x, times(y, z))
   eq(0, 0) -> true
   eq(s(x), 0) -> false
   eq(0, s(y)) -> false
   eq(s(x), s(y)) -> eq(x, y)
   divides(y, x) -> eq(x, times(div(x, y), y))
   prime(s(s(x))) -> pr(s(s(x)), s(x))
   pr(x, s(0)) -> true
   pr(x, s(s(y))) -> if(divides(s(s(y)), x), x, s(y))
   if(true, x, y) -> false
   if(false, x, y) -> pr(x, y)

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

(6) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

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

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

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

(8) 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(x), s(y)) -> EQ(x, y)
The graph contains the following edges 1 > 1, 2 > 2


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

(9)
YES

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

(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

   plus(x, 0) -> x
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   times(0, y) -> 0
   times(s(0), y) -> y
   times(s(x), y) -> plus(y, times(x, y))
   div(0, y) -> 0
   div(x, y) -> quot(x, y, y)
   quot(0, s(y), z) -> 0
   quot(s(x), s(y), z) -> quot(x, y, z)
   quot(x, 0, s(z)) -> s(div(x, s(z)))
   div(div(x, y), z) -> div(x, times(y, z))
   eq(0, 0) -> true
   eq(s(x), 0) -> false
   eq(0, s(y)) -> false
   eq(s(x), s(y)) -> eq(x, y)
   divides(y, x) -> eq(x, times(div(x, y), y))
   prime(s(s(x))) -> pr(s(s(x)), s(x))
   pr(x, s(0)) -> true
   pr(x, s(s(y))) -> if(divides(s(s(y)), x), x, s(y))
   if(true, x, y) -> false
   if(false, x, y) -> pr(x, y)

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

(11) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(12)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

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

(13) 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:
*PLUS(s(x), y) -> PLUS(x, y)
The graph contains the following edges 1 > 1, 2 >= 2


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

(14)
YES

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

(15)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

   plus(x, 0) -> x
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   times(0, y) -> 0
   times(s(0), y) -> y
   times(s(x), y) -> plus(y, times(x, y))
   div(0, y) -> 0
   div(x, y) -> quot(x, y, y)
   quot(0, s(y), z) -> 0
   quot(s(x), s(y), z) -> quot(x, y, z)
   quot(x, 0, s(z)) -> s(div(x, s(z)))
   div(div(x, y), z) -> div(x, times(y, z))
   eq(0, 0) -> true
   eq(s(x), 0) -> false
   eq(0, s(y)) -> false
   eq(s(x), s(y)) -> eq(x, y)
   divides(y, x) -> eq(x, times(div(x, y), y))
   prime(s(s(x))) -> pr(s(s(x)), s(x))
   pr(x, s(0)) -> true
   pr(x, s(s(y))) -> if(divides(s(s(y)), x), x, s(y))
   if(true, x, y) -> false
   if(false, x, y) -> pr(x, y)

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

(16) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(17)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

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

(18) 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:
*TIMES(s(x), y) -> TIMES(x, y)
The graph contains the following edges 1 > 1, 2 >= 2


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

(19)
YES

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

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

   DIV(x, y) -> QUOT(x, y, y)
   QUOT(s(x), s(y), z) -> QUOT(x, y, z)
   QUOT(x, 0, s(z)) -> DIV(x, s(z))
   DIV(div(x, y), z) -> DIV(x, times(y, z))

The TRS R consists of the following rules:

   plus(x, 0) -> x
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   times(0, y) -> 0
   times(s(0), y) -> y
   times(s(x), y) -> plus(y, times(x, y))
   div(0, y) -> 0
   div(x, y) -> quot(x, y, y)
   quot(0, s(y), z) -> 0
   quot(s(x), s(y), z) -> quot(x, y, z)
   quot(x, 0, s(z)) -> s(div(x, s(z)))
   div(div(x, y), z) -> div(x, times(y, z))
   eq(0, 0) -> true
   eq(s(x), 0) -> false
   eq(0, s(y)) -> false
   eq(s(x), s(y)) -> eq(x, y)
   divides(y, x) -> eq(x, times(div(x, y), y))
   prime(s(s(x))) -> pr(s(s(x)), s(x))
   pr(x, s(0)) -> true
   pr(x, s(s(y))) -> if(divides(s(s(y)), x), x, s(y))
   if(true, x, y) -> false
   if(false, x, y) -> pr(x, y)

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

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


The following pairs can be oriented strictly and are deleted.

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

QUOT(x1, x2, x3)  =  x1

s(x1)  =  x1

div(x1, x2)  =  div(x1)


Knuth-Bendix order [KBO] with precedence:trivial

and weight map:

   dummyConstant=1
   div_1=1

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


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

(22)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   DIV(x, y) -> QUOT(x, y, y)
   QUOT(s(x), s(y), z) -> QUOT(x, y, z)
   QUOT(x, 0, s(z)) -> DIV(x, s(z))

The TRS R consists of the following rules:

   plus(x, 0) -> x
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   times(0, y) -> 0
   times(s(0), y) -> y
   times(s(x), y) -> plus(y, times(x, y))
   div(0, y) -> 0
   div(x, y) -> quot(x, y, y)
   quot(0, s(y), z) -> 0
   quot(s(x), s(y), z) -> quot(x, y, z)
   quot(x, 0, s(z)) -> s(div(x, s(z)))
   div(div(x, y), z) -> div(x, times(y, z))
   eq(0, 0) -> true
   eq(s(x), 0) -> false
   eq(0, s(y)) -> false
   eq(s(x), s(y)) -> eq(x, y)
   divides(y, x) -> eq(x, times(div(x, y), y))
   prime(s(s(x))) -> pr(s(s(x)), s(x))
   pr(x, s(0)) -> true
   pr(x, s(s(y))) -> if(divides(s(s(y)), x), x, s(y))
   if(true, x, y) -> false
   if(false, x, y) -> pr(x, y)

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

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

QUOT(x1, x2, x3)  =  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


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

(24)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   DIV(x, y) -> QUOT(x, y, y)
   QUOT(x, 0, s(z)) -> DIV(x, s(z))

The TRS R consists of the following rules:

   plus(x, 0) -> x
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   times(0, y) -> 0
   times(s(0), y) -> y
   times(s(x), y) -> plus(y, times(x, y))
   div(0, y) -> 0
   div(x, y) -> quot(x, y, y)
   quot(0, s(y), z) -> 0
   quot(s(x), s(y), z) -> quot(x, y, z)
   quot(x, 0, s(z)) -> s(div(x, s(z)))
   div(div(x, y), z) -> div(x, times(y, z))
   eq(0, 0) -> true
   eq(s(x), 0) -> false
   eq(0, s(y)) -> false
   eq(s(x), s(y)) -> eq(x, y)
   divides(y, x) -> eq(x, times(div(x, y), y))
   prime(s(s(x))) -> pr(s(s(x)), s(x))
   pr(x, s(0)) -> true
   pr(x, s(s(y))) -> if(divides(s(s(y)), x), x, s(y))
   if(true, x, y) -> false
   if(false, x, y) -> pr(x, y)

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

(25) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule DIV(x, y) -> QUOT(x, y, y) we obtained the following new rules [LPAR04]:

   (DIV(y_0, s(y_1)) -> QUOT(y_0, s(y_1), s(y_1)),DIV(y_0, s(y_1)) -> QUOT(y_0, s(y_1), s(y_1)))


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

(26)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   QUOT(x, 0, s(z)) -> DIV(x, s(z))
   DIV(y_0, s(y_1)) -> QUOT(y_0, s(y_1), s(y_1))

The TRS R consists of the following rules:

   plus(x, 0) -> x
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   times(0, y) -> 0
   times(s(0), y) -> y
   times(s(x), y) -> plus(y, times(x, y))
   div(0, y) -> 0
   div(x, y) -> quot(x, y, y)
   quot(0, s(y), z) -> 0
   quot(s(x), s(y), z) -> quot(x, y, z)
   quot(x, 0, s(z)) -> s(div(x, s(z)))
   div(div(x, y), z) -> div(x, times(y, z))
   eq(0, 0) -> true
   eq(s(x), 0) -> false
   eq(0, s(y)) -> false
   eq(s(x), s(y)) -> eq(x, y)
   divides(y, x) -> eq(x, times(div(x, y), y))
   prime(s(s(x))) -> pr(s(s(x)), s(x))
   pr(x, s(0)) -> true
   pr(x, s(s(y))) -> if(divides(s(s(y)), x), x, s(y))
   if(true, x, y) -> false
   if(false, x, y) -> pr(x, y)

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

(27) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.
----------------------------------------

(28)
TRUE

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

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

   PR(x, s(s(y))) -> IF(divides(s(s(y)), x), x, s(y))
   IF(false, x, y) -> PR(x, y)

The TRS R consists of the following rules:

   plus(x, 0) -> x
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   times(0, y) -> 0
   times(s(0), y) -> y
   times(s(x), y) -> plus(y, times(x, y))
   div(0, y) -> 0
   div(x, y) -> quot(x, y, y)
   quot(0, s(y), z) -> 0
   quot(s(x), s(y), z) -> quot(x, y, z)
   quot(x, 0, s(z)) -> s(div(x, s(z)))
   div(div(x, y), z) -> div(x, times(y, z))
   eq(0, 0) -> true
   eq(s(x), 0) -> false
   eq(0, s(y)) -> false
   eq(s(x), s(y)) -> eq(x, y)
   divides(y, x) -> eq(x, times(div(x, y), y))
   prime(s(s(x))) -> pr(s(s(x)), s(x))
   pr(x, s(0)) -> true
   pr(x, s(s(y))) -> if(divides(s(s(y)), x), x, s(y))
   if(true, x, y) -> false
   if(false, x, y) -> pr(x, y)

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

(30) 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(false, x, y) -> PR(x, y)
The graph contains the following edges 2 >= 1, 3 >= 2


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


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

(31)
YES