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TRS Stand 20472 pair #381709614
details
property
value
status
complete
benchmark
log.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n109.star.cs.uiowa.edu
space
AProVE_08
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
2.5889480114 seconds
cpu usage
7.227755671
max memory
5.38046464E8
stage attributes
key
value
output-size
8666
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 19 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) QDPSizeChangeProof [EQUIVALENT, 0 ms] (9) YES (10) QDP (11) NonMonReductionPairProof [EQUIVALENT, 141 ms] (12) QDP (13) PisEmptyProof [EQUIVALENT, 0 ms] (14) YES (15) QDP (16) NonMonReductionPairProof [EQUIVALENT, 91 ms] (17) QDP (18) PisEmptyProof [EQUIVALENT, 0 ms] (19) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: p(0) -> 0 p(s(x)) -> x minus(x, 0) -> x minus(s(x), s(y)) -> minus(x, y) minus(x, s(y)) -> p(minus(x, y)) div(0, s(y)) -> 0 div(s(x), s(y)) -> s(div(minus(s(x), s(y)), s(y))) log(s(0), s(s(y))) -> 0 log(s(s(x)), s(s(y))) -> s(log(div(minus(x, y), s(s(y))), s(s(y)))) Q is empty. ---------------------------------------- (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) MINUS(x, s(y)) -> P(minus(x, y)) MINUS(x, s(y)) -> MINUS(x, y) DIV(s(x), s(y)) -> DIV(minus(s(x), s(y)), s(y)) DIV(s(x), s(y)) -> MINUS(s(x), s(y)) LOG(s(s(x)), s(s(y))) -> LOG(div(minus(x, y), s(s(y))), s(s(y))) LOG(s(s(x)), s(s(y))) -> DIV(minus(x, y), s(s(y))) LOG(s(s(x)), s(s(y))) -> MINUS(x, y) The TRS R consists of the following rules: p(0) -> 0 p(s(x)) -> x minus(x, 0) -> x minus(s(x), s(y)) -> minus(x, y) minus(x, s(y)) -> p(minus(x, y)) div(0, s(y)) -> 0 div(s(x), s(y)) -> s(div(minus(s(x), s(y)), s(y))) log(s(0), s(s(y))) -> 0 log(s(s(x)), s(s(y))) -> s(log(div(minus(x, y), s(s(y))), s(s(y)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 4 less nodes. ---------------------------------------- (4) Complex Obligation (AND) ---------------------------------------- (5)
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