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SRS_Standard 2019-03-29 03.29 pair #432292429
details
property
value
status
complete
benchmark
z081.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n032.star.cs.uiowa.edu
space
Zantema_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
7.36634 seconds
cpu usage
25.0645
user time
24.1881
system time
0.876357
max virtual memory
3.7906864E7
max residence set size
2574532.0
stage attributes
key
value
starexec-result
YES
output
24.59/7.17 YES 24.71/7.21 proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml 24.71/7.21 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 24.71/7.21 24.71/7.21 24.71/7.21 Termination w.r.t. Q of the given QTRS could be proven: 24.71/7.21 24.71/7.21 (0) QTRS 24.71/7.21 (1) QTRSRRRProof [EQUIVALENT, 53 ms] 24.71/7.21 (2) QTRS 24.71/7.21 (3) DependencyPairsProof [EQUIVALENT, 3 ms] 24.71/7.21 (4) QDP 24.71/7.21 (5) DependencyGraphProof [EQUIVALENT, 0 ms] 24.71/7.21 (6) QDP 24.71/7.21 (7) QDPOrderProof [EQUIVALENT, 18 ms] 24.71/7.21 (8) QDP 24.71/7.21 (9) PisEmptyProof [EQUIVALENT, 0 ms] 24.71/7.21 (10) YES 24.71/7.21 24.71/7.21 24.71/7.21 ---------------------------------------- 24.71/7.21 24.71/7.21 (0) 24.71/7.21 Obligation: 24.71/7.21 Q restricted rewrite system: 24.71/7.21 The TRS R consists of the following rules: 24.71/7.21 24.71/7.21 b(c(a(x1))) -> a(b(x1)) 24.71/7.21 b(b(b(x1))) -> c(a(c(x1))) 24.71/7.21 c(d(x1)) -> d(c(x1)) 24.71/7.21 c(d(b(x1))) -> d(c(c(x1))) 24.71/7.21 d(c(x1)) -> b(b(b(x1))) 24.71/7.21 c(b(x1)) -> d(a(x1)) 24.71/7.21 d(b(c(x1))) -> a(a(x1)) 24.71/7.21 d(a(x1)) -> b(x1) 24.71/7.21 24.71/7.21 Q is empty. 24.71/7.21 24.71/7.21 ---------------------------------------- 24.71/7.21 24.71/7.21 (1) QTRSRRRProof (EQUIVALENT) 24.71/7.21 Used ordering: 24.71/7.21 Polynomial interpretation [POLO]: 24.71/7.21 24.71/7.21 POL(a(x_1)) = x_1 24.71/7.21 POL(b(x_1)) = 1 + x_1 24.71/7.21 POL(c(x_1)) = 1 + x_1 24.71/7.21 POL(d(x_1)) = 2 + x_1 24.71/7.21 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 24.71/7.21 24.71/7.21 b(c(a(x1))) -> a(b(x1)) 24.71/7.21 b(b(b(x1))) -> c(a(c(x1))) 24.71/7.21 d(b(c(x1))) -> a(a(x1)) 24.71/7.21 d(a(x1)) -> b(x1) 24.71/7.21 24.71/7.21 24.71/7.21 24.71/7.21 24.71/7.21 ---------------------------------------- 24.71/7.21 24.71/7.21 (2) 24.71/7.21 Obligation: 24.71/7.21 Q restricted rewrite system: 24.71/7.21 The TRS R consists of the following rules: 24.71/7.21 24.71/7.21 c(d(x1)) -> d(c(x1)) 24.71/7.21 c(d(b(x1))) -> d(c(c(x1))) 24.71/7.21 d(c(x1)) -> b(b(b(x1))) 24.71/7.21 c(b(x1)) -> d(a(x1)) 24.71/7.21 24.71/7.21 Q is empty. 24.71/7.21 24.71/7.21 ---------------------------------------- 24.71/7.21 24.71/7.21 (3) DependencyPairsProof (EQUIVALENT) 24.71/7.21 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 24.71/7.21 ---------------------------------------- 24.71/7.21 24.71/7.21 (4) 24.71/7.21 Obligation: 24.71/7.21 Q DP problem: 24.71/7.21 The TRS P consists of the following rules: 24.71/7.21 24.71/7.21 C(d(x1)) -> D(c(x1)) 24.71/7.21 C(d(x1)) -> C(x1) 24.71/7.21 C(d(b(x1))) -> D(c(c(x1))) 24.71/7.21 C(d(b(x1))) -> C(c(x1)) 24.71/7.21 C(d(b(x1))) -> C(x1) 24.71/7.21 C(b(x1)) -> D(a(x1)) 24.71/7.21 24.71/7.21 The TRS R consists of the following rules: 24.71/7.21 24.71/7.21 c(d(x1)) -> d(c(x1)) 24.71/7.21 c(d(b(x1))) -> d(c(c(x1))) 24.71/7.21 d(c(x1)) -> b(b(b(x1))) 24.71/7.21 c(b(x1)) -> d(a(x1)) 24.71/7.21 24.71/7.21 Q is empty. 24.71/7.21 We have to consider all minimal (P,Q,R)-chains. 24.71/7.21 ----------------------------------------
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