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SRS Stand 10685 pair #381724527
details
property
value
status
complete
benchmark
211978.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n006.star.cs.uiowa.edu
space
ICFP_2010
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
4.33953094482 seconds
cpu usage
14.185803507
max memory
2.187669504E9
stage attributes
key
value
output-size
16559
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 201 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 5 ms] (4) AND (5) QDP (6) QDPSizeChangeProof [EQUIVALENT, 0 ms] (7) YES (8) QDP (9) QDPSizeChangeProof [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(0(1(x1))) -> 0(2(3(0(1(x1))))) 0(0(1(x1))) -> 0(4(0(5(4(1(x1)))))) 0(0(1(x1))) -> 2(1(0(0(3(4(x1)))))) 0(0(1(x1))) -> 4(0(5(4(0(1(x1)))))) 0(1(0(x1))) -> 0(0(2(1(2(x1))))) 0(1(0(x1))) -> 1(0(0(5(4(x1))))) 0(1(0(x1))) -> 0(0(2(5(4(1(x1)))))) 0(1(1(x1))) -> 1(0(3(4(1(x1))))) 0(1(1(x1))) -> 5(0(3(4(1(1(x1)))))) 5(0(1(x1))) -> 0(5(4(1(x1)))) 5(0(1(x1))) -> 2(5(4(0(1(x1))))) 5(0(1(x1))) -> 5(0(2(1(2(x1))))) 5(0(1(x1))) -> 0(1(4(5(4(4(x1)))))) 5(0(1(x1))) -> 0(5(4(1(4(4(x1)))))) 5(0(1(x1))) -> 5(0(4(3(0(1(x1)))))) 5(1(0(x1))) -> 5(0(2(2(1(x1))))) 5(1(0(x1))) -> 5(0(5(4(1(x1))))) 5(1(0(x1))) -> 0(5(0(2(2(1(x1)))))) 5(1(0(x1))) -> 1(4(0(5(2(3(x1)))))) 5(1(0(x1))) -> 1(5(0(4(4(2(x1)))))) 5(1(0(x1))) -> 4(4(1(0(4(5(x1)))))) 5(1(1(x1))) -> 1(1(5(4(x1)))) 5(1(1(x1))) -> 5(4(1(1(x1)))) 5(1(1(x1))) -> 1(5(3(4(1(x1))))) 5(1(1(x1))) -> 1(1(4(5(4(4(x1)))))) 5(1(1(x1))) -> 3(5(2(3(1(1(x1)))))) 5(1(1(x1))) -> 4(1(2(1(5(4(x1)))))) 0(1(3(0(x1)))) -> 0(2(0(2(1(3(x1)))))) 0(1(5(0(x1)))) -> 0(0(5(4(1(5(x1)))))) 0(1(5(0(x1)))) -> 0(5(4(2(1(0(x1)))))) 0(3(0(1(x1)))) -> 0(0(4(1(3(0(x1)))))) 0(3(1(0(x1)))) -> 0(0(2(3(1(x1))))) 0(3(1(1(x1)))) -> 5(1(1(0(3(4(x1)))))) 5(0(1(0(x1)))) -> 5(0(0(4(1(3(x1)))))) 5(1(2(0(x1)))) -> 1(4(0(5(4(2(x1)))))) 5(1(2(0(x1)))) -> 5(0(4(2(2(1(x1)))))) 5(1(4(0(x1)))) -> 1(5(4(0(2(3(x1)))))) 5(1(4(0(x1)))) -> 4(5(2(1(3(0(x1)))))) 5(1(5(1(x1)))) -> 5(4(1(5(1(x1))))) 5(3(0(1(x1)))) -> 0(1(5(2(3(x1))))) 5(3(1(0(x1)))) -> 1(4(3(5(0(x1))))) 5(3(1(0(x1)))) -> 1(5(0(4(3(x1))))) 5(3(1(0(x1)))) -> 5(4(3(1(0(x1))))) 5(3(1(0(x1)))) -> 1(3(0(4(3(5(x1)))))) 5(3(1(1(x1)))) -> 1(1(5(3(3(4(x1)))))) 0(1(2(5(0(x1))))) -> 1(5(4(0(2(0(x1)))))) 0(1(4(2(0(x1))))) -> 1(0(4(2(3(0(x1)))))) 1(4(5(1(0(x1))))) -> 5(4(2(1(1(0(x1)))))) 5(0(1(4(0(x1))))) -> 1(4(5(4(0(0(x1)))))) 5(5(1(0(0(x1))))) -> 5(5(0(4(1(0(x1)))))) Q is empty. ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: 0^1(0(1(x1))) -> 0^1(2(3(0(1(x1))))) 0^1(0(1(x1))) -> 0^1(4(0(5(4(1(x1)))))) 0^1(0(1(x1))) -> 0^1(5(4(1(x1))))
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