details

property	value
status	complete
benchmark	4.30.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n181.star.cs.uiowa.edu
space	SK90

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.98238 seconds
cpu usage	4.7441
user time	4.53718
system time	0.206923
max virtual memory	1.8475704E7
max residence set size	296616.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 68 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 0 ms] (4) QTRS (5) QTRSRRRProof [EQUIVALENT, 8 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) AND (15) QDP (16) UsableRulesProof [EQUIVALENT, 0 ms] (17) QDP (18) QReductionProof [EQUIVALENT, 0 ms] (19) QDP (20) MRRProof [EQUIVALENT, 5 ms] (21) QDP (22) PisEmptyProof [EQUIVALENT, 0 ms] (23) YES (24) QDP (25) UsableRulesProof [EQUIVALENT, 0 ms] (26) QDP (27) QReductionProof [EQUIVALENT, 0 ms] (28) QDP (29) MRRProof [EQUIVALENT, 0 ms] (30) QDP (31) PisEmptyProof [EQUIVALENT, 0 ms] (32) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(nil) -> nil f(.(nil, y)) -> .(nil, f(y)) f(.(.(x, y), z)) -> f(.(x, .(y, z))) g(nil) -> nil g(.(x, nil)) -> .(g(x), nil) g(.(x, .(y, z))) -> g(.(.(x, y), z)) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(.(x_1, x_2)) = x_1 + x_2 POL(f(x_1)) = 2 + x_1 POL(g(x_1)) = x_1 POL(nil) = 0 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: f(nil) -> nil ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(.(nil, y)) -> .(nil, f(y)) f(.(.(x, y), z)) -> f(.(x, .(y, z))) g(nil) -> nil g(.(x, nil)) -> .(g(x), nil) g(.(x, .(y, z))) -> g(.(.(x, y), z)) Q is empty. ---------------------------------------- (3) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(.(x_1, x_2)) = x_1 + x_2 POL(f(x_1)) = 2*x_1 POL(g(x_1)) = x_1 POL(nil) = 1 popout

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all output return to TRS Standard