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SRS Stand 10685 pair #381719697
details
property
value
status
complete
benchmark
160263.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n085.star.cs.uiowa.edu
space
ICFP_2010
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
12.0110940933 seconds
cpu usage
44.274889371
max memory
4.33600512E9
stage attributes
key
value
output-size
26838
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 210 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 179 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 93 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 61 ms] (12) QDP (13) PisEmptyProof [EQUIVALENT, 0 ms] (14) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(1(2(2(x1)))) -> 0(1(0(2(2(x1))))) 0(1(2(2(x1)))) -> 0(1(2(3(2(x1))))) 0(1(2(2(x1)))) -> 0(2(2(1(3(x1))))) 0(1(2(2(x1)))) -> 1(0(3(2(2(x1))))) 0(1(2(2(x1)))) -> 1(2(0(3(2(x1))))) 0(1(2(2(x1)))) -> 1(3(0(2(2(x1))))) 0(1(2(2(x1)))) -> 1(3(2(0(2(x1))))) 0(1(2(2(x1)))) -> 0(1(0(4(2(2(x1)))))) 0(1(2(2(x1)))) -> 0(2(1(3(2(3(x1)))))) 0(1(2(2(x1)))) -> 1(2(1(0(4(2(x1)))))) 0(1(2(2(x1)))) -> 1(5(0(4(2(2(x1)))))) 0(1(2(2(x1)))) -> 2(0(3(1(3(2(x1)))))) 0(1(2(2(x1)))) -> 2(1(1(0(4(2(x1)))))) 0(1(2(2(x1)))) -> 2(1(3(0(2(0(x1)))))) 0(1(2(2(x1)))) -> 2(1(3(3(2(0(x1)))))) 0(1(2(2(x1)))) -> 2(1(5(3(0(2(x1)))))) 0(1(2(2(x1)))) -> 2(2(1(3(0(5(x1)))))) 0(1(2(2(x1)))) -> 2(4(1(3(2(0(x1)))))) 0(1(4(5(x1)))) -> 1(5(0(4(1(x1))))) 0(1(4(5(x1)))) -> 5(0(4(1(5(x1))))) 0(1(4(5(x1)))) -> 5(4(1(5(0(x1))))) 0(1(4(5(x1)))) -> 1(1(5(0(4(1(x1)))))) 0(1(4(5(x1)))) -> 5(4(1(5(5(0(x1)))))) 5(1(2(2(x1)))) -> 1(0(2(2(5(x1))))) 5(1(2(2(x1)))) -> 1(3(5(2(2(x1))))) 5(1(2(2(x1)))) -> 1(5(2(3(2(x1))))) 5(1(2(2(x1)))) -> 1(5(0(2(2(3(x1)))))) 5(1(2(2(x1)))) -> 2(1(0(3(2(5(x1)))))) 5(1(2(2(x1)))) -> 3(1(3(5(2(2(x1)))))) 5(1(2(2(x1)))) -> 4(1(3(2(2(5(x1)))))) 5(1(2(2(x1)))) -> 5(1(0(4(2(2(x1)))))) 5(1(2(2(x1)))) -> 5(1(2(0(4(2(x1)))))) 0(1(1(4(5(x1))))) -> 3(1(0(4(1(5(x1)))))) 0(1(2(2(2(x1))))) -> 1(0(2(2(5(2(x1)))))) 0(1(2(2(5(x1))))) -> 1(5(0(4(2(2(x1)))))) 0(1(2(4(5(x1))))) -> 2(5(1(0(4(5(x1)))))) 0(1(4(5(2(x1))))) -> 1(0(4(2(0(5(x1)))))) 0(1(4(5(5(x1))))) -> 5(0(4(0(1(5(x1)))))) 0(1(5(4(5(x1))))) -> 1(5(0(4(1(5(x1)))))) 0(5(1(2(2(x1))))) -> 0(1(3(2(5(2(x1)))))) 3(3(1(2(2(x1))))) -> 1(3(2(0(3(2(x1)))))) 3(4(4(0(5(x1))))) -> 3(5(4(5(0(4(x1)))))) 5(0(1(2(2(x1))))) -> 1(3(2(0(5(2(x1)))))) 5(1(2(2(5(x1))))) -> 1(5(2(3(2(5(x1)))))) 5(2(1(2(2(x1))))) -> 2(1(3(5(2(2(x1)))))) 5(2(4(0(5(x1))))) -> 0(4(2(5(5(5(x1)))))) 5(2(4(0(5(x1))))) -> 0(4(5(4(2(5(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(1(2(2(x1)))) -> 0^1(1(0(2(2(x1))))) 0^1(1(2(2(x1)))) -> 0^1(2(2(x1)))
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