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SRS Standard pair #487090816
details
property
value
status
complete
benchmark
18.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n142.star.cs.uiowa.edu
space
Gebhardt_06
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
67.5596 seconds
cpu usage
262.579
user time
258.681
system time
3.89811
max virtual memory
5.920072E7
max residence set size
5753672.0
stage attributes
key
value
starexec-result
YES
output
YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) MRRProof [EQUIVALENT, 102 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 462 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(0(0(0(x1)))) -> 1(0(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)))) -> 1^1(0(1(1(x1)))) 0^1(0(0(0(x1)))) -> 0^1(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)))) -> 1(0(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) MRRProof (EQUIVALENT) By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. Strictly oriented dependency pairs: 0^1(0(0(0(x1)))) -> 0^1(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) Used ordering: Polynomial interpretation [POLO]: POL(0(x_1)) = 1 + x_1 POL(0^1(x_1)) = x_1 POL(1(x_1)) = 1 + x_1 POL(1^1(x_1)) = x_1 ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: 0^1(0(0(0(x1)))) -> 1^1(0(1(1(x1)))) 1^1(0(0(1(x1)))) -> 0^1(0(0(0(x1)))) The TRS R consists of the following rules: 0(0(0(0(x1)))) -> 1(0(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. ---------------------------------------- (5) QDPOrderProof (EQUIVALENT)
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