details

property	value
status	complete
benchmark	#4.34.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n142.star.cs.uiowa.edu
space	AG01_innermost

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.87598109245 seconds
cpu usage	3.963139558
max memory	2.44494336E8

stage attributes

key	value
output-size	6687
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 27 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) QReductionProof [EQUIVALENT, 0 ms] (9) QDP (10) QDPSizeChangeProof [EQUIVALENT, 0 ms] (11) YES (12) QDP (13) QDPOrderProof [EQUIVALENT, 0 ms] (14) QDP (15) PisEmptyProof [EQUIVALENT, 0 ms] (16) 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, x, y) -> x if(false, x, y) -> y g(s(x), s(y)) -> if(f(x), s(x), s(y)) g(x, c(y)) -> g(x, g(s(c(y)), y)) The set Q consists of the following terms: f(0) f(1) f(s(x0)) if(true, x0, x1) if(false, x0, x1) g(s(x0), s(x1)) g(x0, c(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(s(x)) -> F(x) G(s(x), s(y)) -> IF(f(x), s(x), s(y)) G(s(x), s(y)) -> F(x) G(x, c(y)) -> G(x, g(s(c(y)), y)) G(x, c(y)) -> G(s(c(y)), y) The TRS R consists of the following rules: f(0) -> true f(1) -> false f(s(x)) -> f(x) if(true, x, y) -> x if(false, x, y) -> y g(s(x), s(y)) -> if(f(x), s(x), s(y)) g(x, c(y)) -> g(x, g(s(c(y)), y)) The set Q consists of the following terms: f(0) f(1) f(s(x0)) if(true, x0, x1) if(false, x0, x1) g(s(x0), s(x1)) g(x0, c(x1)) 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 2 less nodes. popout

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job log

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actions

all output return to TRS Innermost