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TRS Stand 20472 pair #381712568
details
property
value
status
complete
benchmark
Ex49_GM04.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n038.star.cs.uiowa.edu
space
Strategy_removed_CSR_05
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.76609706879 seconds
cpu usage
3.832269931
max memory
1.95244032E8
stage attributes
key
value
output-size
8500
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof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, 14 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) QDPSizeChangeProof [EQUIVALENT, 0 ms] (20) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: minus(0, Y) -> 0 minus(s(X), s(Y)) -> minus(X, Y) geq(X, 0) -> true geq(0, s(Y)) -> false geq(s(X), s(Y)) -> geq(X, Y) div(0, s(Y)) -> 0 div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0) if(true, X, Y) -> X if(false, X, Y) -> 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: minus(0, Y) -> 0 minus(s(X), s(Y)) -> minus(X, Y) geq(X, 0) -> true geq(0, s(Y)) -> false geq(s(X), s(Y)) -> geq(X, Y) div(0, s(Y)) -> 0 div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0) if(true, X, Y) -> X if(false, X, Y) -> Y The set Q consists of the following terms: minus(0, x0) minus(s(x0), s(x1)) geq(x0, 0) geq(0, s(x0)) geq(s(x0), s(x1)) div(0, s(x0)) div(s(x0), s(x1)) if(true, x0, x1) if(false, x0, 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:
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