details

property	value
status	complete
benchmark	#4.26.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n028.star.cs.uiowa.edu
space	AG01_innermost

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.93791389465 seconds
cpu usage	4.534887692
max memory	3.1025152E8

stage attributes

key	value
output-size	8723
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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) QReductionProof [EQUIVALENT, 0 ms] (9) QDP (10) QDPSizeChangeProof [EQUIVALENT, 0 ms] (11) YES (12) QDP (13) UsableRulesProof [EQUIVALENT, 0 ms] (14) QDP (15) QReductionProof [EQUIVALENT, 0 ms] (16) QDP (17) QDPOrderProof [EQUIVALENT, 6 ms] (18) QDP (19) DependencyGraphProof [EQUIVALENT, 0 ms] (20) 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, y) -> if(le(x, y), x, y) if(true, x, y) -> 0 if(false, x, y) -> s(minus(p(x), 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, x1) if(true, x0, x1) if(false, 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: LE(s(x), s(y)) -> LE(x, y) MINUS(x, y) -> IF(le(x, y), x, y) MINUS(x, y) -> LE(x, y) IF(false, x, y) -> MINUS(p(x), y) IF(false, x, y) -> P(x) 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, y) -> if(le(x, y), x, y) if(true, x, y) -> 0 if(false, x, y) -> s(minus(p(x), 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, x1) popout

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job log

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actions

all output return to TRS Inner 89993