details

property	value
status	complete
benchmark	#3.48.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n041.star.cs.uiowa.edu
space	AG01

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.71230006218 seconds
cpu usage	3.683645364
max memory	1.94904064E8

stage attributes

key	value
output-size	5194
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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) QDPSizeChangeProof [EQUIVALENT, 0 ms] (14) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(0) -> true f(1) -> false f(s(x)) -> f(x) if(true, s(x), s(y)) -> s(x) if(false, s(x), s(y)) -> s(y) g(x, c(y)) -> c(g(x, y)) g(x, c(y)) -> g(x, if(f(x), c(g(s(x), y)), c(y))) 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: F(s(x)) -> F(x) G(x, c(y)) -> G(x, y) G(x, c(y)) -> G(x, if(f(x), c(g(s(x), y)), c(y))) G(x, c(y)) -> IF(f(x), c(g(s(x), y)), c(y)) G(x, c(y)) -> F(x) G(x, c(y)) -> G(s(x), y) The TRS R consists of the following rules: f(0) -> true f(1) -> false f(s(x)) -> f(x) if(true, s(x), s(y)) -> s(x) if(false, s(x), s(y)) -> s(y) g(x, c(y)) -> c(g(x, y)) g(x, c(y)) -> g(x, if(f(x), c(g(s(x), y)), c(y))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes. ---------------------------------------- (4) Complex Obligation (AND) ---------------------------------------- (5) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(x)) -> F(x) The TRS R consists of the following rules: f(0) -> true f(1) -> false f(s(x)) -> f(x) popout

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