details

property	value
status	complete
benchmark	22.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n080.star.cs.uiowa.edu
space	Various_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.71906995773 seconds
cpu usage	3.611165252
max memory	1.90316544E8

stage attributes

key	value
output-size	4212
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) AAECC Innermost [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) UsableRulesProof [EQUIVALENT, 0 ms] (8) QDP (9) QReductionProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPSizeChangeProof [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(x, 0) -> s(0) f(s(x), s(y)) -> s(f(x, y)) g(0, x) -> g(f(x, x), x) Q is empty. ---------------------------------------- (1) AAECC Innermost (EQUIVALENT) We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is f(x, 0) -> s(0) f(s(x), s(y)) -> s(f(x, y)) The TRS R 2 is g(0, x) -> g(f(x, x), x) The signature Sigma is {g_2} ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(x, 0) -> s(0) f(s(x), s(y)) -> s(f(x, y)) g(0, x) -> g(f(x, x), x) The set Q consists of the following terms: f(x0, 0) f(s(x0), s(x1)) g(0, x0) ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: F(s(x), s(y)) -> F(x, y) G(0, x) -> G(f(x, x), x) G(0, x) -> F(x, x) The TRS R consists of the following rules: f(x, 0) -> s(0) f(s(x), s(y)) -> s(f(x, y)) g(0, x) -> g(f(x, x), x) The set Q consists of the following terms: f(x0, 0) f(s(x0), s(x1)) g(0, x0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- popout

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