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SRS_Standard 2019-03-29 03.29 pair #432293905
details
property
value
status
complete
benchmark
x01.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n016.star.cs.uiowa.edu
space
Secret_07_SRS
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
11.1534 seconds
cpu usage
39.6533
user time
38.3659
system time
1.2874
max virtual memory
2.0270016E7
max residence set size
4185312.0
stage attributes
key
value
starexec-result
YES
output
21.07/6.38 YES 39.50/11.08 proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml 39.50/11.08 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 39.50/11.08 39.50/11.08 39.50/11.08 Termination w.r.t. Q of the given QTRS could be proven: 39.50/11.08 39.50/11.08 (0) QTRS 39.50/11.08 (1) DependencyPairsProof [EQUIVALENT, 4 ms] 39.50/11.08 (2) QDP 39.50/11.08 (3) QDPOrderProof [EQUIVALENT, 98 ms] 39.50/11.08 (4) QDP 39.50/11.08 (5) DependencyGraphProof [EQUIVALENT, 0 ms] 39.50/11.08 (6) QDP 39.50/11.08 (7) QDPOrderProof [EQUIVALENT, 76 ms] 39.50/11.08 (8) QDP 39.50/11.08 (9) DependencyGraphProof [EQUIVALENT, 0 ms] 39.50/11.08 (10) TRUE 39.50/11.08 39.50/11.08 39.50/11.08 ---------------------------------------- 39.50/11.08 39.50/11.08 (0) 39.50/11.08 Obligation: 39.50/11.08 Q restricted rewrite system: 39.50/11.08 The TRS R consists of the following rules: 39.50/11.08 39.50/11.08 a(x1) -> b(b(x1)) 39.50/11.08 c(b(x1)) -> d(x1) 39.50/11.08 e(b(x1)) -> c(c(x1)) 39.50/11.08 d(b(x1)) -> b(f(x1)) 39.50/11.08 f(x1) -> a(e(x1)) 39.50/11.08 c(x1) -> x1 39.50/11.08 a(a(x1)) -> f(x1) 39.50/11.08 39.50/11.08 Q is empty. 39.50/11.08 39.50/11.08 ---------------------------------------- 39.50/11.08 39.50/11.08 (1) DependencyPairsProof (EQUIVALENT) 39.50/11.08 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 39.50/11.08 ---------------------------------------- 39.50/11.08 39.50/11.08 (2) 39.50/11.08 Obligation: 39.50/11.08 Q DP problem: 39.50/11.08 The TRS P consists of the following rules: 39.50/11.08 39.50/11.08 C(b(x1)) -> D(x1) 39.50/11.08 E(b(x1)) -> C(c(x1)) 39.50/11.08 E(b(x1)) -> C(x1) 39.50/11.08 D(b(x1)) -> F(x1) 39.50/11.08 F(x1) -> A(e(x1)) 39.50/11.08 F(x1) -> E(x1) 39.50/11.08 A(a(x1)) -> F(x1) 39.50/11.08 39.50/11.08 The TRS R consists of the following rules: 39.50/11.08 39.50/11.08 a(x1) -> b(b(x1)) 39.50/11.08 c(b(x1)) -> d(x1) 39.50/11.08 e(b(x1)) -> c(c(x1)) 39.50/11.08 d(b(x1)) -> b(f(x1)) 39.50/11.08 f(x1) -> a(e(x1)) 39.50/11.08 c(x1) -> x1 39.50/11.08 a(a(x1)) -> f(x1) 39.50/11.08 39.50/11.08 Q is empty. 39.50/11.08 We have to consider all minimal (P,Q,R)-chains. 39.50/11.08 ---------------------------------------- 39.50/11.08 39.50/11.08 (3) QDPOrderProof (EQUIVALENT) 39.50/11.08 We use the reduction pair processor [LPAR04,JAR06]. 39.50/11.08 39.50/11.08 39.50/11.08 The following pairs can be oriented strictly and are deleted. 39.50/11.08 39.50/11.08 C(b(x1)) -> D(x1) 39.50/11.08 The remaining pairs can at least be oriented weakly. 39.50/11.08 Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: 39.50/11.08 39.50/11.08 <<< 39.50/11.08 POL(C(x_1)) = [[-I]] + [[0A, 0A, 0A]] * x_1 39.50/11.08 >>> 39.50/11.08 39.50/11.08 <<< 39.50/11.08 POL(b(x_1)) = [[0A], [1A], [0A]] + [[0A, 0A, 0A], [0A, 0A, 1A], [0A, -I, 0A]] * x_1 39.50/11.08 >>> 39.50/11.08 39.50/11.08 <<< 39.50/11.08 POL(D(x_1)) = [[0A]] + [[-I, -I, 0A]] * x_1 39.50/11.08 >>> 39.50/11.08 39.50/11.08 <<< 39.50/11.08 POL(E(x_1)) = [[0A]] + [[0A, -I, 0A]] * x_1 39.50/11.09 >>> 39.50/11.09 39.50/11.09 <<< 39.50/11.09 POL(c(x_1)) = [[-I], [-I], [-I]] + [[0A, 0A, 0A], [0A, 0A, 0A], [0A, 0A, 0A]] * x_1 39.50/11.09 >>> 39.50/11.09
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