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SRS_Standard 2019-03-29 03.29 pair #432289033
details
property
value
status
complete
benchmark
17.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n088.star.cs.uiowa.edu
space
Gebhardt_06
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
6.28911 seconds
cpu usage
21.0204
user time
19.9981
system time
1.02221
max virtual memory
4.1502744E7
max residence set size
2920672.0
stage attributes
key
value
starexec-result
YES
output
20.54/6.19 YES 20.85/6.23 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 20.85/6.23 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 20.85/6.23 20.85/6.23 20.85/6.23 Termination w.r.t. Q of the given QTRS could be proven: 20.85/6.23 20.85/6.23 (0) QTRS 20.85/6.23 (1) QTRS Reverse [EQUIVALENT, 0 ms] 20.85/6.23 (2) QTRS 20.85/6.23 (3) DependencyPairsProof [EQUIVALENT, 21 ms] 20.85/6.23 (4) QDP 20.85/6.23 (5) QDPOrderProof [EQUIVALENT, 68 ms] 20.85/6.23 (6) QDP 20.85/6.23 (7) QDPOrderProof [EQUIVALENT, 77 ms] 20.85/6.23 (8) QDP 20.85/6.23 (9) DependencyGraphProof [EQUIVALENT, 0 ms] 20.85/6.23 (10) TRUE 20.85/6.23 20.85/6.23 20.85/6.23 ---------------------------------------- 20.85/6.23 20.85/6.23 (0) 20.85/6.23 Obligation: 20.85/6.23 Q restricted rewrite system: 20.85/6.23 The TRS R consists of the following rules: 20.85/6.23 20.85/6.23 0(0(0(0(x1)))) -> 1(0(1(1(x1)))) 20.85/6.23 0(1(0(1(x1)))) -> 0(0(1(0(x1)))) 20.85/6.23 20.85/6.23 Q is empty. 20.85/6.23 20.85/6.23 ---------------------------------------- 20.85/6.23 20.85/6.23 (1) QTRS Reverse (EQUIVALENT) 20.85/6.23 We applied the QTRS Reverse Processor [REVERSE]. 20.85/6.23 ---------------------------------------- 20.85/6.23 20.85/6.23 (2) 20.85/6.23 Obligation: 20.85/6.23 Q restricted rewrite system: 20.85/6.23 The TRS R consists of the following rules: 20.85/6.23 20.85/6.23 0(0(0(0(x1)))) -> 1(1(0(1(x1)))) 20.85/6.23 1(0(1(0(x1)))) -> 0(1(0(0(x1)))) 20.85/6.23 20.85/6.23 Q is empty. 20.85/6.23 20.85/6.23 ---------------------------------------- 20.85/6.23 20.85/6.23 (3) DependencyPairsProof (EQUIVALENT) 20.85/6.23 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 20.85/6.23 ---------------------------------------- 20.85/6.23 20.85/6.23 (4) 20.85/6.23 Obligation: 20.85/6.23 Q DP problem: 20.85/6.23 The TRS P consists of the following rules: 20.85/6.23 20.85/6.23 0^1(0(0(0(x1)))) -> 1^1(1(0(1(x1)))) 20.85/6.23 0^1(0(0(0(x1)))) -> 1^1(0(1(x1))) 20.85/6.23 0^1(0(0(0(x1)))) -> 0^1(1(x1)) 20.85/6.23 0^1(0(0(0(x1)))) -> 1^1(x1) 20.85/6.23 1^1(0(1(0(x1)))) -> 0^1(1(0(0(x1)))) 20.85/6.23 1^1(0(1(0(x1)))) -> 1^1(0(0(x1))) 20.85/6.23 1^1(0(1(0(x1)))) -> 0^1(0(x1)) 20.85/6.23 20.85/6.23 The TRS R consists of the following rules: 20.85/6.23 20.85/6.23 0(0(0(0(x1)))) -> 1(1(0(1(x1)))) 20.85/6.23 1(0(1(0(x1)))) -> 0(1(0(0(x1)))) 20.85/6.23 20.85/6.23 Q is empty. 20.85/6.23 We have to consider all minimal (P,Q,R)-chains. 20.85/6.23 ---------------------------------------- 20.85/6.23 20.85/6.23 (5) QDPOrderProof (EQUIVALENT) 20.85/6.23 We use the reduction pair processor [LPAR04,JAR06]. 20.85/6.23 20.85/6.23 20.85/6.23 The following pairs can be oriented strictly and are deleted. 20.85/6.23 20.85/6.23 0^1(0(0(0(x1)))) -> 1^1(0(1(x1))) 20.85/6.23 0^1(0(0(0(x1)))) -> 0^1(1(x1)) 20.85/6.23 0^1(0(0(0(x1)))) -> 1^1(x1) 20.85/6.23 1^1(0(1(0(x1)))) -> 1^1(0(0(x1))) 20.85/6.23 1^1(0(1(0(x1)))) -> 0^1(0(x1)) 20.85/6.23 The remaining pairs can at least be oriented weakly. 20.85/6.23 Used ordering: Polynomial interpretation [POLO]: 20.85/6.23 20.85/6.23 POL(0(x_1)) = 1 + x_1 20.85/6.23 POL(0^1(x_1)) = 1 + x_1 20.85/6.23 POL(1(x_1)) = 1 + x_1 20.85/6.23 POL(1^1(x_1)) = 1 + x_1 20.85/6.23 20.85/6.23 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 20.85/6.23 20.85/6.23 1(0(1(0(x1)))) -> 0(1(0(0(x1)))) 20.85/6.23 0(0(0(0(x1)))) -> 1(1(0(1(x1)))) 20.85/6.23
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