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


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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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) MRRProof [EQUIVALENT, 11 ms]
        (9) QDP
        (10) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (11) YES
    (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, 61 ms]
        (22) QDP
        (23) PisEmptyProof [EQUIVALENT, 0 ms]
        (24) YES


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

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

   minus(minus(x, y), z) -> minus(x, plus(y, z))
   minus(0, y) -> 0
   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   plus(0, y) -> y
   plus(s(x), y) -> plus(x, s(y))
   plus(s(x), y) -> s(plus(y, x))
   zero(s(x)) -> false
   zero(0) -> true
   p(s(x)) -> x
   p(0) -> 0
   div(x, y) -> quot(x, y, 0)
   quot(s(x), s(y), z) -> quot(minus(p(ack(0, x)), y), s(y), s(z))
   quot(0, s(y), z) -> z
   ack(0, x) -> s(x)
   ack(0, x) -> plus(x, s(0))
   ack(s(x), 0) -> ack(x, s(0))
   ack(s(x), s(y)) -> ack(x, ack(s(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:

   MINUS(minus(x, y), z) -> MINUS(x, plus(y, z))
   MINUS(minus(x, y), z) -> PLUS(y, z)
   MINUS(s(x), s(y)) -> MINUS(x, y)
   PLUS(s(x), y) -> PLUS(x, s(y))
   PLUS(s(x), y) -> PLUS(y, x)
   DIV(x, y) -> QUOT(x, y, 0)
   QUOT(s(x), s(y), z) -> QUOT(minus(p(ack(0, x)), y), s(y), s(z))
   QUOT(s(x), s(y), z) -> MINUS(p(ack(0, x)), y)
   QUOT(s(x), s(y), z) -> P(ack(0, x))
   QUOT(s(x), s(y), z) -> ACK(0, x)
   ACK(0, x) -> PLUS(x, s(0))
   ACK(s(x), 0) -> ACK(x, s(0))
   ACK(s(x), s(y)) -> ACK(x, ack(s(x), y))
   ACK(s(x), s(y)) -> ACK(s(x), y)

The TRS R consists of the following rules:

   minus(minus(x, y), z) -> minus(x, plus(y, z))
   minus(0, y) -> 0
   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   plus(0, y) -> y
   plus(s(x), y) -> plus(x, s(y))
   plus(s(x), y) -> s(plus(y, x))
   zero(s(x)) -> false
   zero(0) -> true
   p(s(x)) -> x
   p(0) -> 0
   div(x, y) -> quot(x, y, 0)
   quot(s(x), s(y), z) -> quot(minus(p(ack(0, x)), y), s(y), s(z))
   quot(0, s(y), z) -> z
   ack(0, x) -> s(x)
   ack(0, x) -> plus(x, s(0))
   ack(s(x), 0) -> ack(x, s(0))
   ack(s(x), s(y)) -> ack(x, ack(s(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 4 SCCs with 6 less nodes.
----------------------------------------

(4)
Complex Obligation (AND)

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

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

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

The TRS R consists of the following rules:

   minus(minus(x, y), z) -> minus(x, plus(y, z))
   minus(0, y) -> 0
   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   plus(0, y) -> y
   plus(s(x), y) -> plus(x, s(y))
   plus(s(x), y) -> s(plus(y, x))
   zero(s(x)) -> false
   zero(0) -> true
   p(s(x)) -> x
   p(0) -> 0
   div(x, y) -> quot(x, y, 0)
   quot(s(x), s(y), z) -> quot(minus(p(ack(0, x)), y), s(y), s(z))
   quot(0, s(y), z) -> z
   ack(0, x) -> s(x)
   ack(0, x) -> plus(x, s(0))
   ack(s(x), 0) -> ack(x, s(0))
   ack(s(x), s(y)) -> ack(x, ack(s(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:

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

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

(8) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented dependency pairs:

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


Used ordering: Polynomial interpretation [POLO]:

   POL(PLUS(x_1, x_2)) = 2*x_1 + 2*x_2
   POL(s(x_1)) = 2 + x_1


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

(9)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

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

(10) 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, s(y))
The graph contains the following edges 1 > 1


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

(11)
YES

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

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

   ACK(s(x), s(y)) -> ACK(x, ack(s(x), y))
   ACK(s(x), 0) -> ACK(x, s(0))
   ACK(s(x), s(y)) -> ACK(s(x), y)

The TRS R consists of the following rules:

   minus(minus(x, y), z) -> minus(x, plus(y, z))
   minus(0, y) -> 0
   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   plus(0, y) -> y
   plus(s(x), y) -> plus(x, s(y))
   plus(s(x), y) -> s(plus(y, x))
   zero(s(x)) -> false
   zero(0) -> true
   p(s(x)) -> x
   p(0) -> 0
   div(x, y) -> quot(x, y, 0)
   quot(s(x), s(y), z) -> quot(minus(p(ack(0, x)), y), s(y), s(z))
   quot(0, s(y), z) -> z
   ack(0, x) -> s(x)
   ack(0, x) -> plus(x, s(0))
   ack(s(x), 0) -> ack(x, s(0))
   ack(s(x), s(y)) -> ack(x, ack(s(x), y))

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:
*ACK(s(x), 0) -> ACK(x, s(0))
The graph contains the following edges 1 > 1


*ACK(s(x), s(y)) -> ACK(x, ack(s(x), y))
The graph contains the following edges 1 > 1


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

   MINUS(s(x), s(y)) -> MINUS(x, y)
   MINUS(minus(x, y), z) -> MINUS(x, plus(y, z))

The TRS R consists of the following rules:

   minus(minus(x, y), z) -> minus(x, plus(y, z))
   minus(0, y) -> 0
   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   plus(0, y) -> y
   plus(s(x), y) -> plus(x, s(y))
   plus(s(x), y) -> s(plus(y, x))
   zero(s(x)) -> false
   zero(0) -> true
   p(s(x)) -> x
   p(0) -> 0
   div(x, y) -> quot(x, y, 0)
   quot(s(x), s(y), z) -> quot(minus(p(ack(0, x)), y), s(y), s(z))
   quot(0, s(y), z) -> z
   ack(0, x) -> s(x)
   ack(0, x) -> plus(x, s(0))
   ack(s(x), 0) -> ack(x, s(0))
   ack(s(x), s(y)) -> ack(x, ack(s(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:

   MINUS(s(x), s(y)) -> MINUS(x, y)
   MINUS(minus(x, y), z) -> MINUS(x, plus(y, z))

The TRS R consists of the following rules:

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

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


*MINUS(minus(x, y), z) -> MINUS(x, plus(y, z))
The graph contains the following edges 1 > 1


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

(19)
YES

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

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

   QUOT(s(x), s(y), z) -> QUOT(minus(p(ack(0, x)), y), s(y), s(z))

The TRS R consists of the following rules:

   minus(minus(x, y), z) -> minus(x, plus(y, z))
   minus(0, y) -> 0
   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   plus(0, y) -> y
   plus(s(x), y) -> plus(x, s(y))
   plus(s(x), y) -> s(plus(y, x))
   zero(s(x)) -> false
   zero(0) -> true
   p(s(x)) -> x
   p(0) -> 0
   div(x, y) -> quot(x, y, 0)
   quot(s(x), s(y), z) -> quot(minus(p(ack(0, x)), y), s(y), s(z))
   quot(0, s(y), z) -> z
   ack(0, x) -> s(x)
   ack(0, x) -> plus(x, s(0))
   ack(s(x), 0) -> ack(x, s(0))
   ack(s(x), s(y)) -> ack(x, ack(s(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.

   QUOT(s(x), s(y), z) -> QUOT(minus(p(ack(0, x)), y), s(y), s(z))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( QUOT_3(x_1, ..., x_3) ) = max{0, x_1 - 1}
POL( minus_2(x_1, x_2) ) = x_1
POL( p_1(x_1) ) = max{0, x_1 - 2}
POL( ack_2(x_1, x_2) ) = 2x_1 + x_2 + 2
POL( 0 ) = 0
POL( s_1(x_1) ) = x_1 + 2
POL( plus_2(x_1, x_2) ) = x_1 + x_2

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

   ack(0, x) -> s(x)
   ack(0, x) -> plus(x, s(0))
   p(s(x)) -> x
   p(0) -> 0
   minus(s(x), s(y)) -> minus(x, y)
   minus(minus(x, y), z) -> minus(x, plus(y, z))
   minus(0, y) -> 0
   minus(x, 0) -> x
   plus(0, y) -> y
   plus(s(x), y) -> plus(x, s(y))
   plus(s(x), y) -> s(plus(y, x))


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

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

   minus(minus(x, y), z) -> minus(x, plus(y, z))
   minus(0, y) -> 0
   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   plus(0, y) -> y
   plus(s(x), y) -> plus(x, s(y))
   plus(s(x), y) -> s(plus(y, x))
   zero(s(x)) -> false
   zero(0) -> true
   p(s(x)) -> x
   p(0) -> 0
   div(x, y) -> quot(x, y, 0)
   quot(s(x), s(y), z) -> quot(minus(p(ack(0, x)), y), s(y), s(z))
   quot(0, s(y), z) -> z
   ack(0, x) -> s(x)
   ack(0, x) -> plus(x, s(0))
   ack(s(x), 0) -> ack(x, s(0))
   ack(s(x), s(y)) -> ack(x, ack(s(x), y))

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

(23) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
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(24)
YES