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TRS Stand 20472 pair #381712681
details
property
value
status
complete
benchmark
z05.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n022.star.cs.uiowa.edu
space
Zantema_05
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.72901010513 seconds
cpu usage
3.696673219
max memory
2.40021504E8
stage attributes
key
value
output-size
5471
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) UsableRulesReductionPairsProof [EQUIVALENT, 15 ms] (4) QDP (5) MRRProof [EQUIVALENT, 0 ms] (6) QDP (7) RFCMatchBoundsDPProof [EQUIVALENT, 10 ms] (8) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(a, f(a, x)) -> f(c, f(b, x)) f(b, f(b, x)) -> f(a, f(c, x)) f(c, f(c, x)) -> f(b, f(a, 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: F(a, f(a, x)) -> F(c, f(b, x)) F(a, f(a, x)) -> F(b, x) F(b, f(b, x)) -> F(a, f(c, x)) F(b, f(b, x)) -> F(c, x) F(c, f(c, x)) -> F(b, f(a, x)) F(c, f(c, x)) -> F(a, x) The TRS R consists of the following rules: f(a, f(a, x)) -> f(c, f(b, x)) f(b, f(b, x)) -> f(a, f(c, x)) f(c, f(c, x)) -> f(b, f(a, x)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) UsableRulesReductionPairsProof (EQUIVALENT) First, we A-transformed [FROCOS05] the QDP-Problem. Then we obtain the following A-transformed DP problem. The pairs P are: a1(a(x)) -> c1(b(x)) a1(a(x)) -> b1(x) b1(b(x)) -> a1(c(x)) b1(b(x)) -> c1(x) c1(c(x)) -> b1(a(x)) c1(c(x)) -> a1(x) and the Q and R are: Q restricted rewrite system: The TRS R consists of the following rules: c(c(x)) -> b(a(x)) b(b(x)) -> a(c(x)) a(a(x)) -> c(b(x)) Q is empty. By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well. No dependency pairs are removed. The following rules are removed from R: f(a, f(a, x)) -> f(c, f(b, x)) f(b, f(b, x)) -> f(a, f(c, x)) f(c, f(c, x)) -> f(b, f(a, x)) Used ordering: POLO with Polynomial interpretation [POLO]: POL(a(x_1)) = 2*x_1
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