details

property	value
status	complete
benchmark	#4.27.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n186.star.cs.uiowa.edu
space	Strategy_removed_AG01

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.89117 seconds
cpu usage	3.8735
user time	3.6949
system time	0.178593
max virtual memory	1.8476504E7
max residence set size	248808.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof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, 8 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) QReductionProof [EQUIVALENT, 0 ms] (11) QDP (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) QDP (17) QReductionProof [EQUIVALENT, 0 ms] (18) QDP (19) MRRProof [EQUIVALENT, 0 ms] (20) QDP (21) DependencyGraphProof [EQUIVALENT, 0 ms] (22) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: p(0) -> 0 p(s(x)) -> x le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) minus(x, 0) -> x minus(x, s(y)) -> if(le(x, s(y)), 0, p(minus(x, p(s(y))))) if(true, x, y) -> x if(false, x, y) -> y 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: p(0) -> 0 p(s(x)) -> x le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) minus(x, 0) -> x minus(x, s(y)) -> if(le(x, s(y)), 0, p(minus(x, p(s(y))))) if(true, x, y) -> x if(false, x, y) -> y The set Q consists of the following terms: p(0) p(s(x0)) le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) minus(x0, 0) minus(x0, s(x1)) if(true, x0, x1) if(false, x0, x1) ---------------------------------------- (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: LE(s(x), s(y)) -> LE(x, y) MINUS(x, s(y)) -> IF(le(x, s(y)), 0, p(minus(x, p(s(y))))) MINUS(x, s(y)) -> LE(x, s(y)) MINUS(x, s(y)) -> P(minus(x, p(s(y)))) popout

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all output return to TRS Standard