details

property	value
status	complete
benchmark	Ex4_7_15_Bor03_Z.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n187.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.75571 seconds
cpu usage	3.84994
user time	3.65913
system time	0.190806
max virtual memory	1.8408912E7
max residence set size	253568.0

stage attributes

key	value
starexec-result	YES

output

YES proof of /export/starexec/sandbox/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 62 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 0 ms] (4) QTRS (5) AAECC Innermost [EQUIVALENT, 0 ms] (6) QTRS (7) DependencyPairsProof [EQUIVALENT, 6 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(0) -> cons(0, n__f(s(0))) f(s(0)) -> f(p(s(0))) p(s(0)) -> 0 f(X) -> n__f(X) activate(n__f(X)) -> f(X) 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)) = 1 + 2*x_1 POL(n__f(x_1)) = x_1 POL(p(x_1)) = 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: p(s(0)) -> 0 f(X) -> n__f(X) activate(n__f(X)) -> f(X) activate(X) -> X ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(0) -> cons(0, n__f(s(0))) f(s(0)) -> f(p(s(0))) Q is empty. ---------------------------------------- (3) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(cons(x_1, x_2)) = 2*x_1 + x_2 POL(f(x_1)) = 1 + 2*x_1 POL(n__f(x_1)) = 2*x_1 POL(p(x_1)) = 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: f(0) -> cons(0, n__f(s(0))) ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(s(0)) -> f(p(s(0))) Q is empty. popout

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job log

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actions

all output return to TRS Standard