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SRS_Standard 2019-03-29 03.29 pair #432294967
details
property
value
status
complete
benchmark
random-120.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n154.star.cs.uiowa.edu
space
Waldmann_19
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
3.83112 seconds
cpu usage
11.2439
user time
10.6113
system time
0.632634
max virtual memory
7.835988E7
max residence set size
1833228.0
stage attributes
key
value
starexec-result
YES
output
10.93/3.72 YES 10.93/3.76 proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml 10.93/3.76 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 10.93/3.76 10.93/3.76 10.93/3.76 Termination w.r.t. Q of the given QTRS could be proven: 10.93/3.76 10.93/3.76 (0) QTRS 10.93/3.76 (1) DependencyPairsProof [EQUIVALENT, 17 ms] 10.93/3.76 (2) QDP 10.93/3.76 (3) DependencyGraphProof [EQUIVALENT, 0 ms] 10.93/3.76 (4) QDP 10.93/3.76 (5) MRRProof [EQUIVALENT, 56 ms] 10.93/3.76 (6) QDP 10.93/3.76 (7) PisEmptyProof [EQUIVALENT, 0 ms] 10.93/3.76 (8) YES 10.93/3.76 10.93/3.76 10.93/3.76 ---------------------------------------- 10.93/3.76 10.93/3.76 (0) 10.93/3.76 Obligation: 10.93/3.76 Q restricted rewrite system: 10.93/3.76 The TRS R consists of the following rules: 10.93/3.76 10.93/3.76 b(a(b(a(x1)))) -> b(b(a(b(x1)))) 10.93/3.76 a(a(b(b(x1)))) -> b(b(b(a(x1)))) 10.93/3.76 b(a(b(b(x1)))) -> b(b(a(a(x1)))) 10.93/3.76 10.93/3.76 Q is empty. 10.93/3.76 10.93/3.76 ---------------------------------------- 10.93/3.76 10.93/3.76 (1) DependencyPairsProof (EQUIVALENT) 10.93/3.76 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 10.93/3.76 ---------------------------------------- 10.93/3.76 10.93/3.76 (2) 10.93/3.76 Obligation: 10.93/3.76 Q DP problem: 10.93/3.76 The TRS P consists of the following rules: 10.93/3.76 10.93/3.76 B(a(b(a(x1)))) -> B(b(a(b(x1)))) 10.93/3.76 B(a(b(a(x1)))) -> B(a(b(x1))) 10.93/3.76 B(a(b(a(x1)))) -> A(b(x1)) 10.93/3.76 B(a(b(a(x1)))) -> B(x1) 10.93/3.76 A(a(b(b(x1)))) -> B(b(b(a(x1)))) 10.93/3.76 A(a(b(b(x1)))) -> B(b(a(x1))) 10.93/3.76 A(a(b(b(x1)))) -> B(a(x1)) 10.93/3.76 A(a(b(b(x1)))) -> A(x1) 10.93/3.76 B(a(b(b(x1)))) -> B(b(a(a(x1)))) 10.93/3.76 B(a(b(b(x1)))) -> B(a(a(x1))) 10.93/3.76 B(a(b(b(x1)))) -> A(a(x1)) 10.93/3.76 B(a(b(b(x1)))) -> A(x1) 10.93/3.76 10.93/3.76 The TRS R consists of the following rules: 10.93/3.76 10.93/3.76 b(a(b(a(x1)))) -> b(b(a(b(x1)))) 10.93/3.76 a(a(b(b(x1)))) -> b(b(b(a(x1)))) 10.93/3.76 b(a(b(b(x1)))) -> b(b(a(a(x1)))) 10.93/3.76 10.93/3.76 Q is empty. 10.93/3.76 We have to consider all minimal (P,Q,R)-chains. 10.93/3.76 ---------------------------------------- 10.93/3.76 10.93/3.76 (3) DependencyGraphProof (EQUIVALENT) 10.93/3.76 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 5 less nodes. 10.93/3.76 ---------------------------------------- 10.93/3.76 10.93/3.76 (4) 10.93/3.76 Obligation: 10.93/3.76 Q DP problem: 10.93/3.76 The TRS P consists of the following rules: 10.93/3.76 10.93/3.76 B(a(b(a(x1)))) -> B(x1) 10.93/3.76 B(a(b(a(x1)))) -> B(a(b(x1))) 10.93/3.76 B(a(b(b(x1)))) -> B(a(a(x1))) 10.93/3.76 B(a(b(b(x1)))) -> A(a(x1)) 10.93/3.76 A(a(b(b(x1)))) -> B(a(x1)) 10.93/3.76 B(a(b(b(x1)))) -> A(x1) 10.93/3.76 A(a(b(b(x1)))) -> A(x1) 10.93/3.76 10.93/3.76 The TRS R consists of the following rules: 10.93/3.76 10.93/3.76 b(a(b(a(x1)))) -> b(b(a(b(x1)))) 10.93/3.76 a(a(b(b(x1)))) -> b(b(b(a(x1)))) 10.93/3.76 b(a(b(b(x1)))) -> b(b(a(a(x1)))) 10.93/3.76 10.93/3.76 Q is empty. 10.93/3.76 We have to consider all minimal (P,Q,R)-chains. 10.93/3.76 ---------------------------------------- 10.93/3.76 10.93/3.76 (5) MRRProof (EQUIVALENT) 10.93/3.76 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. 10.93/3.76 10.93/3.76 Strictly oriented dependency pairs: 10.93/3.76 10.93/3.76 B(a(b(a(x1)))) -> B(x1) 10.93/3.76 B(a(b(a(x1)))) -> B(a(b(x1))) 10.93/3.76 B(a(b(b(x1)))) -> B(a(a(x1)))
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