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TRS Stand 20472 pair #381714554
details
property
value
status
complete
benchmark
z15.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n110.star.cs.uiowa.edu
space
Zantema_05
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.65048384666 seconds
cpu usage
3.661031793
max memory
2.27176448E8
stage attributes
key
value
output-size
6165
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) AAECC Innermost [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0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) UsableRulesProof [EQUIVALENT, 0 ms] (6) QDP (7) QReductionProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPToSRSProof [SOUND, 0 ms] (10) QTRS (11) DependencyPairsProof [EQUIVALENT, 0 ms] (12) QDP (13) MNOCProof [EQUIVALENT, 0 ms] (14) QDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) QDP (17) QReductionProof [EQUIVALENT, 0 ms] (18) QDP (19) QDPSizeChangeProof [EQUIVALENT, 1 ms] (20) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(x, a(b(y))) -> f(a(b(x)), y) f(a(x), y) -> f(x, a(y)) f(b(x), y) -> f(x, b(y)) Q is empty. ---------------------------------------- (1) AAECC Innermost (EQUIVALENT) We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is none The TRS R 2 is f(x, a(b(y))) -> f(a(b(x)), y) f(a(x), y) -> f(x, a(y)) f(b(x), y) -> f(x, b(y)) The signature Sigma is {f_2} ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(x, a(b(y))) -> f(a(b(x)), y) f(a(x), y) -> f(x, a(y)) f(b(x), y) -> f(x, b(y)) The set Q consists of the following terms: f(x0, a(b(x1))) f(a(x0), x1) f(b(x0), 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, a(b(y))) -> F(a(b(x)), y) F(a(x), y) -> F(x, a(y)) F(b(x), y) -> F(x, b(y)) The TRS R consists of the following rules: f(x, a(b(y))) -> f(a(b(x)), y) f(a(x), y) -> f(x, a(y)) f(b(x), y) -> f(x, b(y)) The set Q consists of the following terms:
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