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SRS_Standard 2019-03-29 03.29 pair #432292345
details
property
value
status
complete
benchmark
z110.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n012.star.cs.uiowa.edu
space
Zantema_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
4.7312 seconds
cpu usage
14.6644
user time
14.074
system time
0.590344
max virtual memory
5.6203936E7
max residence set size
1695056.0
stage attributes
key
value
starexec-result
YES
output
13.78/4.48 YES 14.26/4.62 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 14.26/4.62 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 14.26/4.62 14.26/4.62 14.26/4.62 Termination w.r.t. Q of the given QTRS could be proven: 14.26/4.62 14.26/4.62 (0) QTRS 14.26/4.62 (1) DependencyPairsProof [EQUIVALENT, 1 ms] 14.26/4.62 (2) QDP 14.26/4.62 (3) MRRProof [EQUIVALENT, 48 ms] 14.26/4.62 (4) QDP 14.26/4.62 (5) DependencyGraphProof [EQUIVALENT, 0 ms] 14.26/4.62 (6) TRUE 14.26/4.62 14.26/4.62 14.26/4.62 ---------------------------------------- 14.26/4.62 14.26/4.62 (0) 14.26/4.62 Obligation: 14.26/4.62 Q restricted rewrite system: 14.26/4.62 The TRS R consists of the following rules: 14.26/4.62 14.26/4.62 a(a(x1)) -> b(x1) 14.26/4.62 b(a(x1)) -> a(b(x1)) 14.26/4.62 b(b(c(x1))) -> c(a(x1)) 14.26/4.62 b(b(x1)) -> a(a(a(x1))) 14.26/4.62 c(a(x1)) -> b(a(c(x1))) 14.26/4.62 14.26/4.62 Q is empty. 14.26/4.62 14.26/4.62 ---------------------------------------- 14.26/4.62 14.26/4.62 (1) DependencyPairsProof (EQUIVALENT) 14.26/4.62 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 14.26/4.62 ---------------------------------------- 14.26/4.62 14.26/4.62 (2) 14.26/4.62 Obligation: 14.26/4.62 Q DP problem: 14.26/4.62 The TRS P consists of the following rules: 14.26/4.62 14.26/4.62 A(a(x1)) -> B(x1) 14.26/4.62 B(a(x1)) -> A(b(x1)) 14.26/4.62 B(a(x1)) -> B(x1) 14.26/4.62 B(b(c(x1))) -> C(a(x1)) 14.26/4.62 B(b(c(x1))) -> A(x1) 14.26/4.62 B(b(x1)) -> A(a(a(x1))) 14.26/4.62 B(b(x1)) -> A(a(x1)) 14.26/4.62 B(b(x1)) -> A(x1) 14.26/4.62 C(a(x1)) -> B(a(c(x1))) 14.26/4.62 C(a(x1)) -> A(c(x1)) 14.26/4.62 C(a(x1)) -> C(x1) 14.26/4.62 14.26/4.62 The TRS R consists of the following rules: 14.26/4.62 14.26/4.62 a(a(x1)) -> b(x1) 14.26/4.62 b(a(x1)) -> a(b(x1)) 14.26/4.62 b(b(c(x1))) -> c(a(x1)) 14.26/4.62 b(b(x1)) -> a(a(a(x1))) 14.26/4.62 c(a(x1)) -> b(a(c(x1))) 14.26/4.62 14.26/4.62 Q is empty. 14.26/4.62 We have to consider all minimal (P,Q,R)-chains. 14.26/4.62 ---------------------------------------- 14.26/4.62 14.26/4.62 (3) MRRProof (EQUIVALENT) 14.26/4.62 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. 14.26/4.62 14.26/4.62 Strictly oriented dependency pairs: 14.26/4.62 14.26/4.62 B(a(x1)) -> A(b(x1)) 14.26/4.62 B(a(x1)) -> B(x1) 14.26/4.62 B(b(c(x1))) -> A(x1) 14.26/4.62 B(b(x1)) -> A(a(a(x1))) 14.26/4.62 B(b(x1)) -> A(a(x1)) 14.26/4.62 B(b(x1)) -> A(x1) 14.26/4.62 C(a(x1)) -> B(a(c(x1))) 14.26/4.62 C(a(x1)) -> A(c(x1)) 14.26/4.62 C(a(x1)) -> C(x1) 14.26/4.62 14.26/4.62 Strictly oriented rules of the TRS R: 14.26/4.62 14.26/4.62 a(a(x1)) -> b(x1) 14.26/4.62 c(a(x1)) -> b(a(c(x1))) 14.26/4.62 14.26/4.62 Used ordering: Polynomial interpretation [POLO]: 14.26/4.62 14.26/4.62 POL(A(x_1)) = 1 + x_1 14.26/4.62 POL(B(x_1)) = 3 + x_1 14.26/4.62 POL(C(x_1)) = 2 + 3*x_1 14.26/4.62 POL(a(x_1)) = 2 + x_1 14.26/4.62 POL(b(x_1)) = 3 + x_1 14.26/4.62 POL(c(x_1)) = 2 + 3*x_1 14.26/4.62 14.26/4.62 14.26/4.62 ---------------------------------------- 14.26/4.62 14.26/4.62 (4) 14.26/4.62 Obligation:
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