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SRS_Standard 2019-03-29 03.29 pair #432295123
details
property
value
status
complete
benchmark
02.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n073.star.cs.uiowa.edu
space
Zantema_06
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
4.49319 seconds
cpu usage
14.2043
user time
13.5271
system time
0.677179
max virtual memory
1.9941564E7
max residence set size
1837344.0
stage attributes
key
value
starexec-result
YES
output
13.29/4.30 YES 13.90/4.45 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 13.90/4.45 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 13.90/4.45 13.90/4.45 13.90/4.45 Termination w.r.t. Q of the given QTRS could be proven: 13.90/4.45 13.90/4.45 (0) QTRS 13.90/4.45 (1) QTRS Reverse [EQUIVALENT, 0 ms] 13.90/4.45 (2) QTRS 13.90/4.45 (3) DependencyPairsProof [EQUIVALENT, 4 ms] 13.90/4.45 (4) QDP 13.90/4.45 (5) MRRProof [EQUIVALENT, 27 ms] 13.90/4.45 (6) QDP 13.90/4.45 (7) DependencyGraphProof [EQUIVALENT, 5 ms] 13.90/4.45 (8) TRUE 13.90/4.45 13.90/4.45 13.90/4.45 ---------------------------------------- 13.90/4.45 13.90/4.45 (0) 13.90/4.45 Obligation: 13.90/4.45 Q restricted rewrite system: 13.90/4.45 The TRS R consists of the following rules: 13.90/4.45 13.90/4.45 a(p(x1)) -> p(a(A(x1))) 13.90/4.45 a(A(x1)) -> A(a(x1)) 13.90/4.45 p(A(A(x1))) -> a(p(x1)) 13.90/4.45 13.90/4.45 Q is empty. 13.90/4.45 13.90/4.45 ---------------------------------------- 13.90/4.45 13.90/4.45 (1) QTRS Reverse (EQUIVALENT) 13.90/4.45 We applied the QTRS Reverse Processor [REVERSE]. 13.90/4.45 ---------------------------------------- 13.90/4.45 13.90/4.45 (2) 13.90/4.45 Obligation: 13.90/4.45 Q restricted rewrite system: 13.90/4.45 The TRS R consists of the following rules: 13.90/4.45 13.90/4.45 p(a(x1)) -> A(a(p(x1))) 13.90/4.45 A(a(x1)) -> a(A(x1)) 13.90/4.45 A(A(p(x1))) -> p(a(x1)) 13.90/4.45 13.90/4.45 Q is empty. 13.90/4.45 13.90/4.45 ---------------------------------------- 13.90/4.45 13.90/4.45 (3) DependencyPairsProof (EQUIVALENT) 13.90/4.45 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 13.90/4.45 ---------------------------------------- 13.90/4.45 13.90/4.45 (4) 13.90/4.45 Obligation: 13.90/4.45 Q DP problem: 13.90/4.45 The TRS P consists of the following rules: 13.90/4.45 13.90/4.45 P(a(x1)) -> A^1(a(p(x1))) 13.90/4.45 P(a(x1)) -> P(x1) 13.90/4.45 A^1(a(x1)) -> A^1(x1) 13.90/4.45 A^1(A(p(x1))) -> P(a(x1)) 13.90/4.45 13.90/4.45 The TRS R consists of the following rules: 13.90/4.45 13.90/4.45 p(a(x1)) -> A(a(p(x1))) 13.90/4.45 A(a(x1)) -> a(A(x1)) 13.90/4.45 A(A(p(x1))) -> p(a(x1)) 13.90/4.45 13.90/4.45 Q is empty. 13.90/4.45 We have to consider all minimal (P,Q,R)-chains. 13.90/4.45 ---------------------------------------- 13.90/4.45 13.90/4.45 (5) MRRProof (EQUIVALENT) 13.90/4.45 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. 13.90/4.45 13.90/4.45 Strictly oriented dependency pairs: 13.90/4.45 13.90/4.45 P(a(x1)) -> P(x1) 13.90/4.45 A^1(a(x1)) -> A^1(x1) 13.90/4.45 A^1(A(p(x1))) -> P(a(x1)) 13.90/4.45 13.90/4.45 Strictly oriented rules of the TRS R: 13.90/4.45 13.90/4.45 A(A(p(x1))) -> p(a(x1)) 13.90/4.45 13.90/4.45 Used ordering: Polynomial interpretation [POLO]: 13.90/4.45 13.90/4.45 POL(A(x_1)) = 2 + x_1 13.90/4.45 POL(A^1(x_1)) = 2 + x_1 13.90/4.45 POL(P(x_1)) = 3*x_1 13.90/4.45 POL(a(x_1)) = 1 + x_1 13.90/4.45 POL(p(x_1)) = 3*x_1 13.90/4.45 13.90/4.45 13.90/4.45 ---------------------------------------- 13.90/4.45 13.90/4.45 (6) 13.90/4.45 Obligation:
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