details

property	value
status	complete
benchmark	Ex1_GM03_Z.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n027.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.13989305496 seconds
cpu usage	4.482240377
max memory	3.09792768E8

stage attributes

key	value
output-size	8924
starexec-result	NO

output

/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- NO 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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 12 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) QDPSizeChangeProof [EQUIVALENT, 0 ms] (9) YES (10) QDP (11) UsableRulesProof [EQUIVALENT, 0 ms] (12) QDP (13) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms] (14) QDP (15) MRRProof [EQUIVALENT, 5 ms] (16) QDP (17) MNOCProof [EQUIVALENT, 0 ms] (18) QDP (19) MNOCProof [EQUIVALENT, 0 ms] (20) QDP (21) NonTerminationLoopProof [COMPLETE, 0 ms] (22) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: p(0) -> 0 p(s(X)) -> X leq(0, Y) -> true leq(s(X), 0) -> false leq(s(X), s(Y)) -> leq(X, Y) if(true, X, Y) -> activate(X) if(false, X, Y) -> activate(Y) diff(X, Y) -> if(leq(X, Y), n__0, n__s(diff(p(X), Y))) 0 -> n__0 s(X) -> n__s(X) activate(n__0) -> 0 activate(n__s(X)) -> s(X) activate(X) -> X 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: LEQ(s(X), s(Y)) -> LEQ(X, Y) IF(true, X, Y) -> ACTIVATE(X) IF(false, X, Y) -> ACTIVATE(Y) DIFF(X, Y) -> IF(leq(X, Y), n__0, n__s(diff(p(X), Y))) DIFF(X, Y) -> LEQ(X, Y) DIFF(X, Y) -> DIFF(p(X), Y) DIFF(X, Y) -> P(X) ACTIVATE(n__0) -> 0^1 ACTIVATE(n__s(X)) -> S(X) The TRS R consists of the following rules: p(0) -> 0 p(s(X)) -> X leq(0, Y) -> true leq(s(X), 0) -> false leq(s(X), s(Y)) -> leq(X, Y) if(true, X, Y) -> activate(X) if(false, X, Y) -> activate(Y) diff(X, Y) -> if(leq(X, Y), n__0, n__s(diff(p(X), Y))) 0 -> n__0 s(X) -> n__s(X) activate(n__0) -> 0 activate(n__s(X)) -> s(X) activate(X) -> X Q is empty. We have to consider all minimal (P,Q,R)-chains. popout

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job log

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actions

all output return to TRS Stand 20472