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TRS Standard pair #516965755
details
property
value
status
complete
benchmark
labeling.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n085.star.cs.uiowa.edu
space
Endrullis_06
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
6.26771402359 seconds
cpu usage
19.470415335
max memory
1.23080704E9
stage attributes
key
value
output-size
60778
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) TransformationProof [EQUIVALENT, 0 ms] (4) QDP (5) TransformationProof [EQUIVALENT, 0 ms] (6) QDP (7) SemLabProof [SOUND, 763 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 301 ms] (12) QDP (13) DependencyGraphProof [EQUIVALENT, 0 ms] (14) QDP (15) QDPOrderProof [EQUIVALENT, 21 ms] (16) QDP (17) QDPOrderProof [EQUIVALENT, 0 ms] (18) QDP (19) PisEmptyProof [EQUIVALENT, 0 ms] (20) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(f(x, y, a), z, w) -> f(z, w, f(y, x, z)) Q is empty. ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: F(f(x, y, a), z, w) -> F(z, w, f(y, x, z)) F(f(x, y, a), z, w) -> F(y, x, z) The TRS R consists of the following rules: f(f(x, y, a), z, w) -> f(z, w, f(y, x, z)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) TransformationProof (EQUIVALENT) By forward instantiating [JAR06] the rule F(f(x, y, a), z, w) -> F(z, w, f(y, x, z)) we obtained the following new rules [LPAR04]: (F(f(x0, x1, a), f(y_0, y_1, a), x3) -> F(f(y_0, y_1, a), x3, f(x1, x0, f(y_0, y_1, a))),F(f(x0, x1, a), f(y_0, y_1, a), x3) -> F(f(y_0, y_1, a), x3, f(x1, x0, f(y_0, y_1, a)))) ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: F(f(x, y, a), z, w) -> F(y, x, z) F(f(x0, x1, a), f(y_0, y_1, a), x3) -> F(f(y_0, y_1, a), x3, f(x1, x0, f(y_0, y_1, a))) The TRS R consists of the following rules: f(f(x, y, a), z, w) -> f(z, w, f(y, x, z)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) TransformationProof (EQUIVALENT) By forward instantiating [JAR06] the rule F(f(x, y, a), z, w) -> F(y, x, z) we obtained the following new rules [LPAR04]: (F(f(x0, f(y_0, y_1, a), a), x2, x3) -> F(f(y_0, y_1, a), x0, x2),F(f(x0, f(y_0, y_1, a), a), x2, x3) -> F(f(y_0, y_1, a), x0, x2)) (F(f(f(y_2, y_3, a), f(y_0, y_1, a), a), x2, x3) -> F(f(y_0, y_1, a), f(y_2, y_3, a), x2),F(f(f(y_2, y_3, a), f(y_0, y_1, a), a), x2, x3) -> F(f(y_0, y_1, a), f(y_2, y_3, a), x2))
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