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TRS Outer 67415 pair #381918228
details
property
value
status
complete
benchmark
test76.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n047.star.cs.uiowa.edu
space
Strategy_outermost_added_08
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
3.61299800873 seconds
cpu usage
11.511247736
max memory
1.3942784E9
stage attributes
key
value
output-size
2141
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Outermost Termination of the given OTRS could be disproven: (0) OTRS (1) OutermostNonTerminationProof [COMPLETE, 1347 ms] (2) NO ---------------------------------------- (0) Obligation: Term rewrite system R: The TRS R consists of the following rules: +(X, 0) -> X +(X, s(Y)) -> s(+(X, Y)) f(0, s(0), X) -> f(X, +(X, X), X) g(X, Y) -> X g(X, Y) -> Y Outermost Strategy. ---------------------------------------- (1) OutermostNonTerminationProof (COMPLETE) Term rewrite system R: The TRS R consists of the following rules: +(X, 0) -> X +(X, s(Y)) -> s(+(X, Y)) f(0, s(0), X) -> f(X, +(X, X), X) g(X, Y) -> X g(X, Y) -> Y Outermost Strategy. ---------- Loop: ---------- f(g(0, s(0)), +(g(0, s(0)), g(0, s(0))), g(0, s(0))) -> f(g(0, s(0)), +(s(0), g(0, s(0))), g(0, s(0))) with rule g(X', Y'') -> Y'' at position [1,0] and matcher [X' / 0, Y'' / s(0)] f(g(0, s(0)), +(s(0), g(0, s(0))), g(0, s(0))) -> f(g(0, s(0)), +(s(0), 0), g(0, s(0))) with rule g(X', Y') -> X' at position [1,1] and matcher [X' / 0, Y' / s(0)] f(g(0, s(0)), +(s(0), 0), g(0, s(0))) -> f(g(0, s(0)), s(0), g(0, s(0))) with rule +(X', 0) -> X' at position [1] and matcher [X' / s(0)] f(g(0, s(0)), s(0), g(0, s(0))) -> f(0, s(0), g(0, s(0))) with rule g(X', Y) -> X' at position [0] and matcher [X' / 0, Y / s(0)] f(0, s(0), g(0, s(0))) -> f(g(0, s(0)), +(g(0, s(0)), g(0, s(0))), g(0, s(0))) with rule f(0, s(0), X) -> f(X, +(X, X), X) at position [] and matcher [X / g(0, s(0))] Now an instance of the first term with Matcher [ ] occurs in the last term at position []. Context: [] We used [THIEMANN_LOOPS_UNDER_STRATEGIES] to show that this Loop is an Outermost-Loop. ---------------------------------------- (2) NO
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