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TRS Stand 20472 pair #381710961
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
n111.star.cs.uiowa.edu
space
AProVE_10
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
12.4636068344 seconds
cpu usage
21.783435982
max memory
5.72461056E8
stage attributes
key
value
output-size
8611
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) AAECC Innermost [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT,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, 81 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))
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