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SRS Stand 10685 pair #381719700
details
property
value
status
complete
benchmark
z114.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n047.star.cs.uiowa.edu
space
Zantema_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
2.82625985146 seconds
cpu usage
7.711773428
max memory
6.23460352E8
stage attributes
key
value
output-size
3418
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES 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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) MRRProof [EQUIVALENT, 38 ms] (4) QDP (5) MRRProof [EQUIVALENT, 13 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(c(x1)) -> c(b(x1)) a(x1) -> b(b(b(x1))) b(c(b(x1))) -> a(c(x1)) 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: A(c(x1)) -> B(x1) A(x1) -> B(b(b(x1))) A(x1) -> B(b(x1)) A(x1) -> B(x1) B(c(b(x1))) -> A(c(x1)) The TRS R consists of the following rules: a(c(x1)) -> c(b(x1)) a(x1) -> b(b(b(x1))) b(c(b(x1))) -> a(c(x1)) 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: A(c(x1)) -> B(x1) Used ordering: Polynomial interpretation [POLO]: POL(A(x_1)) = x_1 POL(B(x_1)) = x_1 POL(a(x_1)) = x_1 POL(b(x_1)) = x_1 POL(c(x_1)) = 1 + x_1 ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: A(x1) -> B(b(b(x1))) A(x1) -> B(b(x1)) A(x1) -> B(x1) B(c(b(x1))) -> A(c(x1)) The TRS R consists of the following rules: a(c(x1)) -> c(b(x1)) a(x1) -> b(b(b(x1))) b(c(b(x1))) -> a(c(x1)) Q is empty.
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