details

property	value
status	complete
benchmark	#4.27.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n070.star.cs.uiowa.edu
space	AG01_innermost

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.7373239994 seconds
cpu usage	3.790884753
max memory	2.44371456E8

stage attributes

key	value
output-size	8211
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 10 ms] (2) QDP (3) DependencyGraphProof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) MRRProof [EQUIVALENT, 0 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, 0) -> x minus(x, s(y)) -> if(le(x, s(y)), 0, p(minus(x, p(s(y))))) if(true, x, y) -> x if(false, x, y) -> 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, 0) minus(x0, s(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, s(y)) -> IF(le(x, s(y)), 0, p(minus(x, p(s(y))))) MINUS(x, s(y)) -> LE(x, s(y)) MINUS(x, s(y)) -> P(minus(x, p(s(y)))) MINUS(x, s(y)) -> MINUS(x, p(s(y))) MINUS(x, s(y)) -> P(s(y)) 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, 0) -> x minus(x, s(y)) -> if(le(x, s(y)), 0, p(minus(x, p(s(y))))) if(true, x, y) -> x if(false, x, y) -> y The set Q consists of the following terms: p(0) p(s(x0)) popout

output may be truncated. 'popout' for the full output.

job log

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actions

all output return to TRS Inner 89993