details

property	value
status	complete
benchmark	ex1.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n173.star.cs.uiowa.edu
space	AProVE_10

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	12.457212925 seconds
cpu usage	27.542262632
max memory	8.32708608E8

stage attributes

key	value
output-size	8605
starexec-result	NO

output

/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- NO proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) AAECC Innermost [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) QReductionProof [EQUIVALENT, 0 ms] (11) QDP (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) MNOCProof [EQUIVALENT, 0 ms] (16) QDP (17) NonLoopProof [COMPLETE, 112 ms] (18) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(x, y) -> cond(lt(x, y), x, y) cond(tt, x, y) -> f(s(x), s(y)) lt(0, y) -> tt lt(s(x), s(y)) -> lt(x, y) Q is empty. ---------------------------------------- (1) AAECC Innermost (EQUIVALENT) We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is lt(0, y) -> tt lt(s(x), s(y)) -> lt(x, y) The TRS R 2 is f(x, y) -> cond(lt(x, y), x, y) cond(tt, x, y) -> f(s(x), s(y)) The signature Sigma is {f_2, cond_3} ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(x, y) -> cond(lt(x, y), x, y) cond(tt, x, y) -> f(s(x), s(y)) lt(0, y) -> tt lt(s(x), s(y)) -> lt(x, y) The set Q consists of the following terms: f(x0, x1) cond(tt, x0, x1) lt(0, x0) lt(s(x0), s(x1)) ---------------------------------------- (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(x, y) -> COND(lt(x, y), x, y) F(x, y) -> LT(x, y) COND(tt, x, y) -> F(s(x), s(y)) LT(s(x), s(y)) -> LT(x, y) The TRS R consists of the following rules: f(x, y) -> cond(lt(x, y), x, y) cond(tt, x, y) -> f(s(x), s(y)) popout

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

job log

popout

actions

all output return to TRS Standard