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	n185.star.cs.uiowa.edu
space	Mixed_TRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.75177 seconds
cpu usage	3.84643
user time	3.67619
system time	0.170243
max virtual memory	1.8343148E7
max residence set size	243268.0

stage attributes

key	value
starexec-result	YES

output

YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 1 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) 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) popout

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all output return to TRS Standard