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SRS Stand 10685 pair #381718156
details
property
value
status
complete
benchmark
09.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n005.star.cs.uiowa.edu
space
Gebhardt_06
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
60.6495680809 seconds
cpu usage
237.939601004
max memory
6.151426048E9
stage attributes
key
value
output-size
4799
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) QDPOrderProof [EQUIVALENT, 39 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 5 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 418 ms] (8) QDP (9) PisEmptyProof [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(0(0(0(x1)))) -> 0(1(1(1(x1)))) 1(0(0(1(x1)))) -> 0(0(0(0(x1)))) 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: 0^1(0(0(0(x1)))) -> 0^1(1(1(1(x1)))) 0^1(0(0(0(x1)))) -> 1^1(1(1(x1))) 0^1(0(0(0(x1)))) -> 1^1(1(x1)) 0^1(0(0(0(x1)))) -> 1^1(x1) 1^1(0(0(1(x1)))) -> 0^1(0(0(0(x1)))) 1^1(0(0(1(x1)))) -> 0^1(0(0(x1))) 1^1(0(0(1(x1)))) -> 0^1(0(x1)) 1^1(0(0(1(x1)))) -> 0^1(x1) The TRS R consists of the following rules: 0(0(0(0(x1)))) -> 0(1(1(1(x1)))) 1(0(0(1(x1)))) -> 0(0(0(0(x1)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. 0^1(0(0(0(x1)))) -> 1^1(1(1(x1))) 0^1(0(0(0(x1)))) -> 1^1(1(x1)) 0^1(0(0(0(x1)))) -> 1^1(x1) 1^1(0(0(1(x1)))) -> 0^1(0(0(x1))) 1^1(0(0(1(x1)))) -> 0^1(0(x1)) 1^1(0(0(1(x1)))) -> 0^1(x1) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial interpretation [POLO]: POL(0(x_1)) = 1 + x_1 POL(0^1(x_1)) = 1 + x_1 POL(1(x_1)) = 1 + x_1 POL(1^1(x_1)) = 1 + x_1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 1(0(0(1(x1)))) -> 0(0(0(0(x1)))) 0(0(0(0(x1)))) -> 0(1(1(1(x1)))) ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules:
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