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TRS Stand 20472 pair #381712786
details
property
value
status
complete
benchmark
cime4.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n079.star.cs.uiowa.edu
space
Secret_05_TRS
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.85265779495 seconds
cpu usage
4.193926902
max memory
3.00761088E8
stage attributes
key
value
output-size
3578
starexec-result
NO
output
/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- NO 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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) NonTerminationLoopProof [COMPLETE, 0 ms] (6) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: g(X) -> u(h(X), h(X), X) u(d, c(Y), X) -> k(Y) h(d) -> c(a) h(d) -> c(b) f(k(a), k(b), X) -> f(X, X, 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: G(X) -> U(h(X), h(X), X) G(X) -> H(X) F(k(a), k(b), X) -> F(X, X, X) The TRS R consists of the following rules: g(X) -> u(h(X), h(X), X) u(d, c(Y), X) -> k(Y) h(d) -> c(a) h(d) -> c(b) f(k(a), k(b), X) -> f(X, X, X) 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 1 SCC with 2 less nodes. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: F(k(a), k(b), X) -> F(X, X, X) The TRS R consists of the following rules: g(X) -> u(h(X), h(X), X) u(d, c(Y), X) -> k(Y) h(d) -> c(a) h(d) -> c(b) f(k(a), k(b), X) -> f(X, X, X) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) NonTerminationLoopProof (COMPLETE) We used the non-termination processor [FROCOS05] to show that the DP problem is infinite. Found a loop by narrowing to the left: s = F(u(d, h(d), X'), u(d, h(d), X''), X) evaluates to t =F(X, X, X) Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * Matcher: [ ] * Semiunifier: [X'' / X', X / u(d, h(d), X')] -------------------------------------------------------------------------------- Rewriting sequence
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