details

property	value
status	complete
benchmark	Ex4_7_15_Bor03_FR.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n055.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.77557396889 seconds
cpu usage	4.070431906
max memory	2.52440576E8

stage attributes

key	value
output-size	5217
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 52 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 12 ms] (4) QTRS (5) QTRSRRRProof [EQUIVALENT, 0 ms] (6) QTRS (7) QTRSRRRProof [EQUIVALENT, 0 ms] (8) QTRS (9) AAECC Innermost [EQUIVALENT, 0 ms] (10) QTRS (11) DependencyPairsProof [EQUIVALENT, 0 ms] (12) QDP (13) DependencyGraphProof [EQUIVALENT, 0 ms] (14) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(0) -> cons(0, n__f(n__s(n__0))) f(s(0)) -> f(p(s(0))) p(s(0)) -> 0 f(X) -> n__f(X) s(X) -> n__s(X) 0 -> n__0 activate(n__f(X)) -> f(activate(X)) activate(n__s(X)) -> s(activate(X)) activate(n__0) -> 0 activate(X) -> X Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 1 POL(activate(x_1)) = 2 + 2*x_1 POL(cons(x_1, x_2)) = x_1 + x_2 POL(f(x_1)) = 2 + 2*x_1 POL(n__0) = 0 POL(n__f(x_1)) = 2 + 2*x_1 POL(n__s(x_1)) = x_1 POL(p(x_1)) = x_1 POL(s(x_1)) = x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: f(0) -> cons(0, n__f(n__s(n__0))) 0 -> n__0 activate(n__0) -> 0 activate(X) -> X ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(s(0)) -> f(p(s(0))) p(s(0)) -> 0 f(X) -> n__f(X) s(X) -> n__s(X) activate(n__f(X)) -> f(activate(X)) activate(n__s(X)) -> s(activate(X)) Q is empty. ---------------------------------------- (3) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 1 POL(activate(x_1)) = 2 + 2*x_1 POL(f(x_1)) = 2 + 2*x_1 popout

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