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TRS Stand 20472 pair #381709812
details
property
value
status
complete
benchmark
test9.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n041.star.cs.uiowa.edu
space
Strategy_removed_mixed_05
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.90557122231 seconds
cpu usage
4.232632965
max memory
3.02280704E8
stage attributes
key
value
output-size
2499
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) NonTerminationLoopProof [COMPLETE, 40 ms] (4) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(0, 1, X) -> h(X, X) h(0, X) -> f(0, 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) -> H(X, X) H(0, X) -> F(0, X, X) The TRS R consists of the following rules: f(0, 1, X) -> h(X, X) h(0, X) -> f(0, X, X) g(X, Y) -> X g(X, Y) -> Y Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) 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 = H(g(X, 0), g(1, Y)) evaluates to t =H(g(1, Y), g(1, Y)) Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * Matcher: [ ] * Semiunifier: [X / 1, Y / 0] -------------------------------------------------------------------------------- Rewriting sequence H(g(1, 0), g(1, 0)) -> H(0, g(1, 0)) with rule g(X, Y') -> Y' at position [0] and matcher [X / 1, Y' / 0] H(0, g(1, 0)) -> F(0, g(1, 0), g(1, 0)) with rule H(0, X') -> F(0, X', X') at position [] and matcher [X' / g(1, 0)] F(0, g(1, 0), g(1, 0)) -> F(0, 1, g(1, 0)) with rule g(X', Y) -> X' at position [1] and matcher [X' / 1, Y / 0] F(0, 1, g(1, 0)) -> H(g(1, 0), g(1, 0)) with rule F(0, 1, X) -> H(X, X) Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence All these steps are and every following step will be a correct step w.r.t to Q. ---------------------------------------- (4) NO
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