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TRS Stand 20472 pair #381709963
details
property
value
status
complete
benchmark
pair3hard.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n086.star.cs.uiowa.edu
space
Endrullis_06
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
3.23078393936 seconds
cpu usage
8.931392779
max memory
6.01698304E8
stage attributes
key
value
output-size
17335
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- 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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) QDPOrderProof [EQUIVALENT, 0 ms] (4) QDP (5) TransformationProof [EQUIVALENT, 0 ms] (6) QDP (7) SemLabProof [SOUND, 127 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) AND (11) QDP (12) QDPOrderProof [EQUIVALENT, 5 ms] (13) QDP (14) PisEmptyProof [EQUIVALENT, 0 ms] (15) YES (16) QDP (17) QDPOrderProof [EQUIVALENT, 23 ms] (18) QDP (19) PisEmptyProof [EQUIVALENT, 0 ms] (20) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: p(a(x0), p(x1, p(x2, x3))) -> p(x1, p(x0, p(a(x3), x3))) 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: P(a(x0), p(x1, p(x2, x3))) -> P(x1, p(x0, p(a(x3), x3))) P(a(x0), p(x1, p(x2, x3))) -> P(x0, p(a(x3), x3)) P(a(x0), p(x1, p(x2, x3))) -> P(a(x3), x3) The TRS R consists of the following rules: p(a(x0), p(x1, p(x2, x3))) -> p(x1, p(x0, p(a(x3), x3))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. P(a(x0), p(x1, p(x2, x3))) -> P(x0, p(a(x3), x3)) P(a(x0), p(x1, p(x2, x3))) -> P(a(x3), x3) The remaining pairs can at least be oriented weakly. Used ordering: Combined order from the following AFS and order. P(x1, x2) = x2 p(x1, x2) = p(x2) Knuth-Bendix order [KBO] with precedence:trivial and weight map: dummyConstant=1 p_1=1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: p(a(x0), p(x1, p(x2, x3))) -> p(x1, p(x0, p(a(x3), x3))) ---------------------------------------- (4)
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