/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) QDPSizeChangeProof [EQUIVALENT, 0 ms]
(4) YES


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

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

Q is empty.

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(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   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)
   F(s(x), y) -> F(x, s(x))
   F(x, s(y)) -> F(y, x)
   F(x, y) -> ACK(x, y)
   ACK(s(x), y) -> F(x, x)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(3) 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), s(y)) -> ACK(s(x), y)
The graph contains the following edges 1 >= 1, 2 > 2


*F(x, y) -> ACK(x, y)
The graph contains the following edges 1 >= 1, 2 >= 2


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


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


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


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


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


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(4)
YES