TRS Inner 89993 pair #381733499

details

property	value
status	complete
benchmark	test77.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n040.star.cs.uiowa.edu
space	Mixed_innermost

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.03400206566 seconds
cpu usage	4.381199677
max memory	2.58080768E8

stage attributes

key	value
output-size	9421
starexec-result	YES

output

/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) QReductionProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (11) YES
    (12) QDP
        (13) UsableRulesProof [EQUIVALENT, 0 ms]
        (14) QDP
        (15) QReductionProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) TransformationProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) UsableRulesProof [EQUIVALENT, 0 ms]
        (20) QDP
        (21) QReductionProof [EQUIVALENT, 0 ms]
        (22) QDP
        (23) TransformationProof [EQUIVALENT, 0 ms]
        (24) QDP
        (25) DependencyGraphProof [EQUIVALENT, 0 ms]
        (26) TRUE


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

   +(X, 0) -> X
   +(X, s(Y)) -> s(+(X, Y))
   double(X) -> +(X, X)
   f(0, s(0), X) -> f(X, double(X), X)
   g(X, Y) -> X
   g(X, Y) -> Y

The set Q consists of the following terms:

   +(x0, 0)
   +(x0, s(x1))
   double(x0)
   f(0, s(0), x0)
   g(x0, x1)


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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:

   +^1(X, s(Y)) -> +^1(X, Y)
   DOUBLE(X) -> +^1(X, X)
   F(0, s(0), X) -> F(X, double(X), X)
   F(0, s(0), X) -> DOUBLE(X)

The TRS R consists of the following rules:

   +(X, 0) -> X
   +(X, s(Y)) -> s(+(X, Y))
   double(X) -> +(X, X)
   f(0, s(0), X) -> f(X, double(X), X)
   g(X, Y) -> X
   g(X, Y) -> Y

The set Q consists of the following terms:

   +(x0, 0)
   +(x0, s(x1))
   double(x0)
   f(0, s(0), x0)
   g(x0, x1)

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