details

property	value
status	complete
benchmark	Ex4_7_15_Bor03_GM.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n095.star.cs.uiowa.edu
space	Transformed_CSR_innermost_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.58539390564 seconds
cpu usage	4.555920409
max memory	2.21679616E8

stage attributes

key	value
output-size	8897
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 83 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 0 ms] (4) QTRS (5) QTRSRRRProof [EQUIVALENT, 11 ms] (6) QTRS (7) QTRSRRRProof [EQUIVALENT, 0 ms] (8) QTRS (9) QTRSRRRProof [EQUIVALENT, 0 ms] (10) QTRS (11) DependencyPairsProof [EQUIVALENT, 0 ms] (12) QDP (13) DependencyGraphProof [EQUIVALENT, 0 ms] (14) QDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) QDP (17) QReductionProof [EQUIVALENT, 0 ms] (18) QDP (19) QDPSizeChangeProof [EQUIVALENT, 0 ms] (20) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__f(0) -> cons(0, f(s(0))) a__f(s(0)) -> a__f(a__p(s(0))) a__p(s(0)) -> 0 mark(f(X)) -> a__f(mark(X)) mark(p(X)) -> a__p(mark(X)) mark(0) -> 0 mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(s(X)) -> s(mark(X)) a__f(X) -> f(X) a__p(X) -> p(X) The set Q consists of the following terms: mark(f(x0)) mark(p(x0)) mark(0) mark(cons(x0, x1)) mark(s(x0)) a__f(x0) a__p(x0) ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(a__f(x_1)) = 2 + 2*x_1 POL(a__p(x_1)) = 2*x_1 POL(cons(x_1, x_2)) = x_1 + x_2 POL(f(x_1)) = 1 + 2*x_1 POL(mark(x_1)) = 2*x_1 POL(p(x_1)) = 2*x_1 POL(s(x_1)) = 2*x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: a__f(0) -> cons(0, f(s(0))) a__f(X) -> f(X) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__f(s(0)) -> a__f(a__p(s(0))) a__p(s(0)) -> 0 mark(f(X)) -> a__f(mark(X)) mark(p(X)) -> a__p(mark(X)) mark(0) -> 0 mark(cons(X1, X2)) -> cons(mark(X1), X2) popout

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job log

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actions

all output return to TRS Inner 89993