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TRS Stand 20472 pair #381713332
details
property
value
status
complete
benchmark
#3.2.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n045.star.cs.uiowa.edu
space
AG01
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.72943902016 seconds
cpu usage
3.779939227
max memory
2.33889792E8
stage attributes
key
value
output-size
8343
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) QDP (8) UsableRulesProof [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) QReductionProof [EQUIVALENT, 0 ms] (18) QDP (19) QDPOrderProof [EQUIVALENT, 0 ms] (20) QDP (21) PisEmptyProof [EQUIVALENT, 0 ms] (22) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: pred(s(x)) -> x minus(x, 0) -> x minus(x, s(y)) -> pred(minus(x, y)) quot(0, s(y)) -> 0 quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) Q is empty. ---------------------------------------- (1) Overlay + Local Confluence (EQUIVALENT) The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: pred(s(x)) -> x minus(x, 0) -> x minus(x, s(y)) -> pred(minus(x, y)) quot(0, s(y)) -> 0 quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) The set Q consists of the following terms: pred(s(x0)) minus(x0, 0) minus(x0, s(x1)) quot(0, s(x0)) quot(s(x0), s(x1)) ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: MINUS(x, s(y)) -> PRED(minus(x, y)) MINUS(x, s(y)) -> MINUS(x, y) QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y)) QUOT(s(x), s(y)) -> MINUS(x, y) The TRS R consists of the following rules: pred(s(x)) -> x minus(x, 0) -> x minus(x, s(y)) -> pred(minus(x, y))
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