details

property	value
status	complete
benchmark	perfect.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n006.star.cs.uiowa.edu
space	Mixed_TRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.7073469162 seconds
cpu usage	4.008645134
max memory	2.38440448E8

stage attributes

key	value
output-size	5676
starexec-result	YES

output

/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 11 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) UsableRulesProof [EQUIVALENT, 0 ms] (8) QDP (9) QReductionProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPSizeChangeProof [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: perfectp(0) -> false perfectp(s(x)) -> f(x, s(0), s(x), s(x)) f(0, y, 0, u) -> true f(0, y, s(z), u) -> false f(s(x), 0, z, u) -> f(x, u, minus(z, s(x)), u) f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)) Q is empty. ---------------------------------------- (1) Overlay + Local Confluence (EQUIVALENT) The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: perfectp(0) -> false perfectp(s(x)) -> f(x, s(0), s(x), s(x)) f(0, y, 0, u) -> true f(0, y, s(z), u) -> false f(s(x), 0, z, u) -> f(x, u, minus(z, s(x)), u) f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)) The set Q consists of the following terms: perfectp(0) perfectp(s(x0)) f(0, x0, 0, x1) f(0, x0, s(x1), x2) f(s(x0), 0, x1, x2) f(s(x0), s(x1), x2, x3) ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: PERFECTP(s(x)) -> F(x, s(0), s(x), s(x)) F(s(x), 0, z, u) -> F(x, u, minus(z, s(x)), u) F(s(x), s(y), z, u) -> F(s(x), minus(y, x), z, u) F(s(x), s(y), z, u) -> F(x, u, z, u) The TRS R consists of the following rules: perfectp(0) -> false perfectp(s(x)) -> f(x, s(0), s(x), s(x)) f(0, y, 0, u) -> true f(0, y, s(z), u) -> false f(s(x), 0, z, u) -> f(x, u, minus(z, s(x)), u) f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)) The set Q consists of the following terms: perfectp(0) popout

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