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TRS Stand 20472 pair #381710473
details
property
value
status
complete
benchmark
pair2hard.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n003.star.cs.uiowa.edu
space
Endrullis_06
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
39.7519009113 seconds
cpu usage
100.044956458
max memory
5.523181568E9
stage attributes
key
value
output-size
7921
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms] (9) QDP (10) PisEmptyProof [EQUIVALENT, 0 ms] (11) YES (12) QDP (13) QDPOrderProof [EQUIVALENT, 0 ms] (14) QDP (15) TransformationProof [EQUIVALENT, 0 ms] (16) QDP (17) QDPOrderProof [EQUIVALENT, 19.7 s] (18) QDP (19) PisEmptyProof [EQUIVALENT, 0 ms] (20) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: p(a(x0), p(b(a(x1)), x2)) -> p(x1, p(a(b(a(x1))), x2)) a(b(a(x0))) -> b(a(b(x0))) 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: P(a(x0), p(b(a(x1)), x2)) -> P(x1, p(a(b(a(x1))), x2)) P(a(x0), p(b(a(x1)), x2)) -> P(a(b(a(x1))), x2) P(a(x0), p(b(a(x1)), x2)) -> A(b(a(x1))) A(b(a(x0))) -> A(b(x0)) The TRS R consists of the following rules: p(a(x0), p(b(a(x1)), x2)) -> p(x1, p(a(b(a(x1))), x2)) a(b(a(x0))) -> b(a(b(x0))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node. ---------------------------------------- (4) Complex Obligation (AND) ---------------------------------------- (5) Obligation: Q DP problem: The TRS P consists of the following rules: A(b(a(x0))) -> A(b(x0)) The TRS R consists of the following rules: p(a(x0), p(b(a(x1)), x2)) -> p(x1, p(a(b(a(x1))), x2)) a(b(a(x0))) -> b(a(b(x0))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (6) UsableRulesProof (EQUIVALENT) We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
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