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SRS Stand 10685 pair #381717603
details
property
value
status
complete
benchmark
dj.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n080.star.cs.uiowa.edu
space
Secret_07_SRS
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
5.0219039917 seconds
cpu usage
16.713741431
max memory
2.22529536E9
stage attributes
key
value
output-size
3568
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 37 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 25 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 1 ms] (6) QDP (7) UsableRulesProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPSizeChangeProof [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 1(0(x1)) -> 0(0(0(1(x1)))) 0(1(x1)) -> 1(x1) 1(1(x1)) -> 0(0(0(0(x1)))) 0(0(x1)) -> 0(x1) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0(x_1)) = x_1 POL(1(x_1)) = 1 + x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 1(1(x1)) -> 0(0(0(0(x1)))) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 1(0(x1)) -> 0(0(0(1(x1)))) 0(1(x1)) -> 1(x1) 0(0(x1)) -> 0(x1) Q is empty. ---------------------------------------- (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: 1^1(0(x1)) -> 0^1(0(0(1(x1)))) 1^1(0(x1)) -> 0^1(0(1(x1))) 1^1(0(x1)) -> 0^1(1(x1)) 1^1(0(x1)) -> 1^1(x1) The TRS R consists of the following rules: 1(0(x1)) -> 0(0(0(1(x1)))) 0(1(x1)) -> 1(x1) 0(0(x1)) -> 0(x1) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes. ---------------------------------------- (6) Obligation:
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