details

property	value
status	complete
benchmark	test75.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n138.star.cs.uiowa.edu
space	Mixed_innermost

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.42911 seconds
cpu usage	4.65696
user time	4.43249
system time	0.224463
max virtual memory	3.6708732E7
max residence set size	323420.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) UsableRulesProof [EQUIVALENT, 0 ms] (6) QDP (7) QReductionProof [EQUIVALENT, 1 ms] (8) QDP (9) TransformationProof [EQUIVALENT, 0 ms] (10) QDP (11) Induction-Processor [SOUND, 32 ms] (12) AND (13) QDP (14) PisEmptyProof [EQUIVALENT, 0 ms] (15) YES (16) QTRS (17) QTRSRRRProof [EQUIVALENT, 0 ms] (18) QTRS (19) RisEmptyProof [EQUIVALENT, 0 ms] (20) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(0, 1, X) -> f(g(X, X), X, X) g(X, Y) -> X g(X, Y) -> Y The set Q consists of the following terms: f(0, 1, x0) g(x0, x1) ---------------------------------------- (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: F(0, 1, X) -> F(g(X, X), X, X) F(0, 1, X) -> G(X, X) The TRS R consists of the following rules: f(0, 1, X) -> f(g(X, X), X, X) g(X, Y) -> X g(X, Y) -> Y The set Q consists of the following terms: f(0, 1, x0) g(x0, x1) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: F(0, 1, X) -> F(g(X, X), X, X) The TRS R consists of the following rules: f(0, 1, X) -> f(g(X, X), X, X) g(X, Y) -> X g(X, Y) -> Y The set Q consists of the following terms: f(0, 1, x0) g(x0, x1) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- popout

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actions

all output return to TRS Innermost