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SRS Standard pair #487083148
details
property
value
status
complete
benchmark
dj.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n139.star.cs.uiowa.edu
space
Secret_07_SRS
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
3.30983 seconds
cpu usage
10.0906
user time
9.61179
system time
0.47877
max virtual memory
2.0340516E7
max residence set size
1243588.0
stage attributes
key
value
starexec-result
YES
output
YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 39 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 10 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) UsableRulesProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPSizeChangeProof [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 1(0(x1)) -> 0(0(0(1(x1)))) 0(1(x1)) -> 1(x1) 1(1(x1)) -> 0(0(0(0(x1)))) 0(0(x1)) -> 0(x1) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0(x_1)) = x_1 POL(1(x_1)) = 1 + x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 1(1(x1)) -> 0(0(0(0(x1)))) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 1(0(x1)) -> 0(0(0(1(x1)))) 0(1(x1)) -> 1(x1) 0(0(x1)) -> 0(x1) Q is empty. ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: 1^1(0(x1)) -> 0^1(0(0(1(x1)))) 1^1(0(x1)) -> 0^1(0(1(x1))) 1^1(0(x1)) -> 0^1(1(x1)) 1^1(0(x1)) -> 1^1(x1) The TRS R consists of the following rules: 1(0(x1)) -> 0(0(0(1(x1)))) 0(1(x1)) -> 1(x1) 0(0(x1)) -> 0(x1) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: 1^1(0(x1)) -> 1^1(x1) The TRS R consists of the following rules:
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