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TRS Stand 20472 pair #381710151
details
property
value
status
complete
benchmark
z22.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n058.star.cs.uiowa.edu
space
Zantema_05
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
19.2616078854 seconds
cpu usage
42.430478493
max memory
4.289142784E9
stage attributes
key
value
output-size
18370
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 13 ms] (2) QDP (3) MRRProof [EQUIVALENT, 36 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 32 ms] (6) QDP (7) SemLabProof [SOUND, 226 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) QDP (11) MRRProof [EQUIVALENT, 8 ms] (12) QDP (13) MRRProof [EQUIVALENT, 5 ms] (14) QDP (15) QDPOrderProof [EQUIVALENT, 10 ms] (16) QDP (17) DependencyGraphProof [EQUIVALENT, 0 ms] (18) AND (19) QDP (20) QDPOrderProof [EQUIVALENT, 6 ms] (21) QDP (22) PisEmptyProof [EQUIVALENT, 0 ms] (23) YES (24) QDP (25) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms] (26) QDP (27) PisEmptyProof [EQUIVALENT, 0 ms] (28) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(f(f(a, b), c), x) -> f(b, f(a, f(c, f(b, x)))) f(x, f(y, z)) -> f(f(x, y), 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(f(a, b), c), x) -> F(b, f(a, f(c, f(b, x)))) F(f(f(a, b), c), x) -> F(a, f(c, f(b, x))) F(f(f(a, b), c), x) -> F(c, f(b, x)) F(f(f(a, b), c), x) -> F(b, x) F(x, f(y, z)) -> F(f(x, y), z) F(x, f(y, z)) -> F(x, y) The TRS R consists of the following rules: f(f(f(a, b), c), x) -> f(b, f(a, f(c, f(b, x)))) f(x, f(y, z)) -> f(f(x, y), z) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) MRRProof (EQUIVALENT) By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. Strictly oriented dependency pairs: F(f(f(a, b), c), x) -> F(c, f(b, x)) F(f(f(a, b), c), x) -> F(b, x) Used ordering: Polynomial interpretation [POLO]: POL(F(x_1, x_2)) = 2*x_1 + 2*x_2 POL(a) = 1 POL(b) = 0 POL(c) = 0 POL(f(x_1, x_2)) = x_1 + x_2
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