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	n036.star.cs.uiowa.edu
space	SK90

run statistics

property	value
solver	AProVE23
configuration	standard
runtime (wallclock)	1.71193099022 seconds
cpu usage	4.333595191
max memory	2.59252224E8

stage attributes

key	value
output-size	4851
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# AProVE Commit ID: 363435ff5b19173ec8a83b45cfdec073678dabcc jan-christoph 20230815 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) AAECC Innermost [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms]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) QDPOrderProof [EQUIVALENT, 4 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