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TRS Stand 20472 pair #381709632
details
property
value
status
complete
benchmark
#4.4.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n080.star.cs.uiowa.edu
space
Strategy_removed_AG01
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.90815091133 seconds
cpu usage
4.119854466
max memory
2.50236928E8
stage attributes
key
value
output-size
2849
starexec-result
NO
output
/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- NO 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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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: f(g(x), x, y) -> f(y, y, g(y)) g(g(x)) -> g(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(g(x), x, y) -> F(y, y, g(y)) F(g(x), x, y) -> G(y) The TRS R consists of the following rules: f(g(x), x, y) -> f(y, y, g(y)) g(g(x)) -> g(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 1 less node. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: F(g(x), x, y) -> F(y, y, g(y)) The TRS R consists of the following rules: f(g(x), x, y) -> f(y, y, g(y)) g(g(x)) -> g(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(g(g(x')), g(g(x')), y) evaluates to t =F(y, y, g(y)) Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * Matcher: [x' / x'', x'' / g(x'')] * Semiunifier: [y / g(g(x''))] -------------------------------------------------------------------------------- Rewriting sequence F(g(g(x')), g(g(x')), g(g(x''))) -> F(g(g(x')), g(x'), g(g(x''))) with rule g(g(x''')) -> g(x''') at position [1] and matcher [x''' / x'] F(g(g(x')), g(x'), g(g(x''))) -> F(g(g(x'')), g(g(x'')), g(g(g(x'')))) with rule F(g(x), x, y) -> F(y, y, g(y)) 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.
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