details

property	value
status	complete
benchmark	4.41.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n067.star.cs.uiowa.edu
space	SK90

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	2.1854300499 seconds
cpu usage	5.713459735
max memory	4.73747456E8

stage attributes

key	value
output-size	4778
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) AAECC Innermost [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 4 ms]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) QDPOrderProof [EQUIVALENT, 0 ms] (12) QDP (13) PisEmptyProof [EQUIVALENT, 0 ms] (14) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(a, y) -> f(y, g(y)) g(a) -> b g(b) -> b Q is empty. ---------------------------------------- (1) AAECC Innermost (EQUIVALENT) We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is g(a) -> b g(b) -> b The TRS R 2 is f(a, y) -> f(y, g(y)) The signature Sigma is {f_2} ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(a, y) -> f(y, g(y)) g(a) -> b g(b) -> b The set Q consists of the following terms: f(a, x0) g(a) g(b) ---------------------------------------- (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(a, y) -> F(y, g(y)) F(a, y) -> G(y) The TRS R consists of the following rules: f(a, y) -> f(y, g(y)) g(a) -> b g(b) -> b The set Q consists of the following terms: f(a, x0) g(a) g(b) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- popout

output may be truncated. 'popout' for the full output.

job log

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actions

all output return to TRS Standard