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TRS Innermost pair #516979593
details
property
value
status
complete
benchmark
#4.30a.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n136.star.cs.uiowa.edu
space
AG01_innermost
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
2.58009004593 seconds
cpu usage
6.824037525
max memory
4.93051904E8
stage attributes
key
value
output-size
15918
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) QReductionProof [EQUIVALENT, 0 ms] (9) QDP (10) QDPSizeChangeProof [EQUIVALENT, 0 ms] (11) YES (12) QDP (13) UsableRulesProof [EQUIVALENT, 0 ms] (14) QDP (15) QReductionProof [EQUIVALENT, 0 ms] (16) QDP (17) QDPSizeChangeProof [EQUIVALENT, 0 ms] (18) YES (19) QDP (20) UsableRulesProof [EQUIVALENT, 0 ms] (21) QDP (22) QReductionProof [EQUIVALENT, 0 ms] (23) QDP (24) Induction-Processor [SOUND, 67 ms] (25) AND (26) QDP (27) PisEmptyProof [EQUIVALENT, 0 ms] (28) YES (29) QTRS (30) QTRSRRRProof [EQUIVALENT, 0 ms] (31) QTRS (32) RisEmptyProof [EQUIVALENT, 0 ms] (33) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: minus(x, 0) -> x minus(s(x), s(y)) -> minus(x, y) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) quot(0, s(y)) -> 0 quot(s(x), s(y)) -> s(quot(minus(s(x), s(y)), s(y))) The set Q consists of the following terms: minus(x0, 0) minus(s(x0), s(x1)) le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) quot(0, s(x0)) quot(s(x0), s(x1)) ---------------------------------------- (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: MINUS(s(x), s(y)) -> MINUS(x, y) LE(s(x), s(y)) -> LE(x, y) QUOT(s(x), s(y)) -> QUOT(minus(s(x), s(y)), s(y)) QUOT(s(x), s(y)) -> MINUS(s(x), s(y)) The TRS R consists of the following rules: minus(x, 0) -> x minus(s(x), s(y)) -> minus(x, y) le(0, y) -> true le(s(x), 0) -> false le(s(x), s(y)) -> le(x, y) quot(0, s(y)) -> 0 quot(s(x), s(y)) -> s(quot(minus(s(x), s(y)), s(y)))
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