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TRS Inner 89993 pair #381733438
details
property
value
status
complete
benchmark
#4.19.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n029.star.cs.uiowa.edu
space
Applicative_AG01_innermost
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.91108489037 seconds
cpu usage
4.528935359
max memory
2.69672448E8
stage attributes
key
value
output-size
17709
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES 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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 1 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) ATransformationProof [EQUIVALENT, 0 ms] (9) QDP (10) QReductionProof [EQUIVALENT, 0 ms] (11) QDP (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) QDP (17) ATransformationProof [EQUIVALENT, 0 ms] (18) QDP (19) QReductionProof [EQUIVALENT, 0 ms] (20) QDP (21) QDPSizeChangeProof [EQUIVALENT, 0 ms] (22) YES (23) QDP (24) QDPSizeChangeProof [EQUIVALENT, 0 ms] (25) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: app(app(app(f, x), app(c, x)), app(c, y)) -> app(app(app(f, y), y), app(app(app(f, y), x), y)) app(app(app(f, app(s, x)), y), z) -> app(app(app(f, x), app(s, app(c, y))), app(c, z)) app(app(app(f, app(c, x)), x), y) -> app(c, y) app(app(g, x), y) -> x app(app(g, x), y) -> y app(app(map, fun), nil) -> nil app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs)) app(app(filter, fun), nil) -> nil app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs) app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs)) app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs) The set Q consists of the following terms: app(app(app(f, x0), app(c, x0)), app(c, x1)) app(app(app(f, app(s, x0)), x1), x2) app(app(app(f, app(c, x0)), x0), x1) app(app(g, x0), x1) app(app(map, x0), nil) app(app(map, x0), app(app(cons, x1), x2)) app(app(filter, x0), nil) app(app(filter, x0), app(app(cons, x1), x2)) app(app(app(app(filter2, true), x0), x1), x2) app(app(app(app(filter2, false), x0), x1), x2) ---------------------------------------- (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: APP(app(app(f, x), app(c, x)), app(c, y)) -> APP(app(app(f, y), y), app(app(app(f, y), x), y)) APP(app(app(f, x), app(c, x)), app(c, y)) -> APP(app(f, y), y) APP(app(app(f, x), app(c, x)), app(c, y)) -> APP(f, y) APP(app(app(f, x), app(c, x)), app(c, y)) -> APP(app(app(f, y), x), y) APP(app(app(f, x), app(c, x)), app(c, y)) -> APP(app(f, y), x) APP(app(app(f, app(s, x)), y), z) -> APP(app(app(f, x), app(s, app(c, y))), app(c, z)) APP(app(app(f, app(s, x)), y), z) -> APP(app(f, x), app(s, app(c, y))) APP(app(app(f, app(s, x)), y), z) -> APP(f, x) APP(app(app(f, app(s, x)), y), z) -> APP(s, app(c, y)) APP(app(app(f, app(s, x)), y), z) -> APP(c, y) APP(app(app(f, app(s, x)), y), z) -> APP(c, z) APP(app(app(f, app(c, x)), x), y) -> APP(c, y) APP(app(map, fun), app(app(cons, x), xs)) -> APP(app(cons, app(fun, x)), app(app(map, fun), xs)) APP(app(map, fun), app(app(cons, x), xs)) -> APP(cons, app(fun, x)) APP(app(map, fun), app(app(cons, x), xs)) -> APP(fun, x)
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