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TRS Stand 20472 pair #381712881
details
property
value
status
complete
benchmark
2.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n099.star.cs.uiowa.edu
space
Secret_06_TRS
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
15.2169630527 seconds
cpu usage
38.038400386
max memory
3.181715456E9
stage attributes
key
value
output-size
6780
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) MRRProof [EQUIVALENT, 14 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) TransformationProof [EQUIVALENT, 0 ms] (8) QDP (9) TransformationProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 2670 ms] (12) QDP (13) DependencyGraphProof [EQUIVALENT, 0 ms] (14) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: c(c(c(b(x)))) -> a(1, b(c(x))) b(c(b(c(x)))) -> a(0, a(1, x)) a(0, x) -> c(c(x)) a(1, x) -> c(b(x)) 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: C(c(c(b(x)))) -> A(1, b(c(x))) C(c(c(b(x)))) -> B(c(x)) C(c(c(b(x)))) -> C(x) B(c(b(c(x)))) -> A(0, a(1, x)) B(c(b(c(x)))) -> A(1, x) A(0, x) -> C(c(x)) A(0, x) -> C(x) A(1, x) -> C(b(x)) A(1, x) -> B(x) The TRS R consists of the following rules: c(c(c(b(x)))) -> a(1, b(c(x))) b(c(b(c(x)))) -> a(0, a(1, x)) a(0, x) -> c(c(x)) a(1, x) -> c(b(x)) 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: C(c(c(b(x)))) -> B(c(x)) C(c(c(b(x)))) -> C(x) B(c(b(c(x)))) -> A(1, x) A(0, x) -> C(x) A(1, x) -> B(x) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(1) = 0 POL(A(x_1, x_2)) = 1 + x_1 + x_2 POL(B(x_1)) = x_1 POL(C(x_1)) = x_1 POL(a(x_1, x_2)) = 2 + 2*x_1 + x_2 POL(b(x_1)) = 1 + x_1 POL(c(x_1)) = 1 + x_1
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