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TRS Standard pair #516964420
details
property
value
status
complete
benchmark
test75.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
Strategy_removed_mixed_05
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
2.46229290962 seconds
cpu usage
6.486679246
max memory
4.95284224E8
stage attributes
key
value
output-size
3163
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 4 ms] (4) QDP (5) NonTerminationLoopProof [COMPLETE, 0 ms] (6) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(0, 1, X) -> f(g(X, X), X, X) g(X, Y) -> X g(X, Y) -> Y 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(0, 1, X) -> F(g(X, X), X, X) F(0, 1, X) -> G(X, X) The TRS R consists of the following rules: f(0, 1, X) -> f(g(X, X), X, X) g(X, Y) -> X g(X, Y) -> Y 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 1 less node. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: F(0, 1, X) -> F(g(X, X), X, X) The TRS R consists of the following rules: f(0, 1, X) -> f(g(X, X), X, X) g(X, Y) -> X g(X, Y) -> Y 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(g(g(0, Y), Y'), g(X', 1), X) evaluates to t =F(g(X, X), X, X) Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * Matcher: [ ] * Semiunifier: [Y / 1, Y' / g(0, 1), X' / 0, X / g(0, 1)] -------------------------------------------------------------------------------- Rewriting sequence F(g(g(0, 1), g(0, 1)), g(0, 1), g(0, 1)) -> F(g(g(0, 1), g(0, 1)), 1, g(0, 1)) with rule g(X', Y'') -> Y'' at position [1] and matcher [X' / 0, Y'' / 1] F(g(g(0, 1), g(0, 1)), 1, g(0, 1)) -> F(g(0, 1), 1, g(0, 1)) with rule g(X', Y') -> X' at position [0] and matcher [X' / g(0, 1), Y' / g(0, 1)] F(g(0, 1), 1, g(0, 1)) -> F(0, 1, g(0, 1))
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