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TRS Outer 67415 pair #381918158
details
property
value
status
complete
benchmark
Ex5_7_Luc97.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n078.star.cs.uiowa.edu
space
Strategy_outermost_added_08
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.27036380768 seconds
cpu usage
3.020258478
max memory
1.97369856E8
stage attributes
key
value
output-size
2288
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Outermost Termination of the given OTRS could be disproven: (0) OTRS (1) OutermostNonTerminationProof [COMPLETE, 45 ms] (2) NO ---------------------------------------- (0) Obligation: Term rewrite system R: The TRS R consists of the following rules: dbl(0) -> 0 dbl(s(X)) -> s(s(dbl(X))) dbls(nil) -> nil dbls(cons(X, Y)) -> cons(dbl(X), dbls(Y)) sel(0, cons(X, Y)) -> X sel(s(X), cons(Y, Z)) -> sel(X, Z) indx(nil, X) -> nil indx(cons(X, Y), Z) -> cons(sel(X, Z), indx(Y, Z)) from(X) -> cons(X, from(s(X))) dbl1(0) -> 01 dbl1(s(X)) -> s1(s1(dbl1(X))) sel1(0, cons(X, Y)) -> X sel1(s(X), cons(Y, Z)) -> sel1(X, Z) quote(0) -> 01 quote(s(X)) -> s1(quote(X)) quote(dbl(X)) -> dbl1(X) quote(sel(X, Y)) -> sel1(X, Y) Outermost Strategy. ---------------------------------------- (1) OutermostNonTerminationProof (COMPLETE) Term rewrite system R: The TRS R consists of the following rules: dbl(0) -> 0 dbl(s(X)) -> s(s(dbl(X))) dbls(nil) -> nil dbls(cons(X, Y)) -> cons(dbl(X), dbls(Y)) sel(0, cons(X, Y)) -> X sel(s(X), cons(Y, Z)) -> sel(X, Z) indx(nil, X) -> nil indx(cons(X, Y), Z) -> cons(sel(X, Z), indx(Y, Z)) from(X) -> cons(X, from(s(X))) dbl1(0) -> 01 dbl1(s(X)) -> s1(s1(dbl1(X))) sel1(0, cons(X, Y)) -> X sel1(s(X), cons(Y, Z)) -> sel1(X, Z) quote(0) -> 01 quote(s(X)) -> s1(quote(X)) quote(dbl(X)) -> dbl1(X) quote(sel(X, Y)) -> sel1(X, Y) Outermost Strategy. ---------- Loop: ---------- from(X) -> cons(X, from(s(X))) with rule from(X') -> cons(X', from(s(X'))) at position [] and matcher [X' / X] Now an instance of the first term with Matcher [X / s(X)] occurs in the last term at position [1]. Context: cons(X, []) We used [THIEMANN_LOOPS_UNDER_STRATEGIES] to show that this Loop is an Outermost-Loop. ---------------------------------------- (2) NO
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