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SRS Standard pair #516970792
details
property
value
status
complete
benchmark
213281.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n077.star.cs.uiowa.edu
space
ICFP_2010
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
6.78586912155 seconds
cpu usage
23.551509432
max memory
1.32970496E9
stage attributes
key
value
output-size
20762
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 200 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 3 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 230 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) QDP (9) UsableRulesProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPSizeChangeProof [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(1(0(2(x1)))) -> 0(0(3(1(2(x1))))) 0(1(3(4(x1)))) -> 0(4(1(0(3(x1))))) 0(1(3(4(x1)))) -> 0(4(1(1(3(x1))))) 0(1(3(4(x1)))) -> 0(4(1(3(1(x1))))) 0(2(1(4(x1)))) -> 0(4(1(2(3(x1))))) 0(2(1(4(x1)))) -> 0(4(1(3(2(x1))))) 0(2(1(4(x1)))) -> 2(0(4(1(4(x1))))) 0(2(1(4(x1)))) -> 5(5(0(4(1(2(x1)))))) 0(2(1(5(x1)))) -> 5(0(4(1(2(x1))))) 0(2(2(4(x1)))) -> 0(4(2(2(5(x1))))) 0(2(2(4(x1)))) -> 0(4(2(5(2(x1))))) 3(4(0(2(x1)))) -> 3(0(4(5(2(x1))))) 3(4(0(2(x1)))) -> 3(5(0(4(2(x1))))) 0(0(1(4(5(x1))))) -> 0(4(1(0(3(5(x1)))))) 0(1(0(2(4(x1))))) -> 2(0(0(4(1(1(x1)))))) 0(1(2(3(4(x1))))) -> 2(0(4(1(0(3(x1)))))) 0(1(3(3(4(x1))))) -> 0(0(3(1(3(4(x1)))))) 0(1(4(0(2(x1))))) -> 0(4(1(5(0(2(x1)))))) 0(1(4(1(5(x1))))) -> 2(5(0(4(1(1(x1)))))) 0(1(4(3(4(x1))))) -> 0(4(0(3(1(4(x1)))))) 0(1(4(3(4(x1))))) -> 3(0(4(1(5(4(x1)))))) 0(1(4(3(5(x1))))) -> 5(4(5(0(3(1(x1)))))) 0(1(5(0(2(x1))))) -> 0(0(4(1(2(5(x1)))))) 0(1(5(1(4(x1))))) -> 4(5(0(3(1(1(x1)))))) 0(2(1(4(4(x1))))) -> 0(4(1(2(4(3(x1)))))) 0(2(1(4(5(x1))))) -> 0(4(1(2(5(2(x1)))))) 0(2(1(5(4(x1))))) -> 5(0(2(0(4(1(x1)))))) 0(2(4(1(5(x1))))) -> 5(0(4(1(5(2(x1)))))) 0(2(4(3(5(x1))))) -> 0(4(5(2(5(3(x1)))))) 0(2(5(1(4(x1))))) -> 0(0(5(4(1(2(x1)))))) 3(0(1(3(2(x1))))) -> 0(3(1(0(3(2(x1)))))) 3(0(2(1(4(x1))))) -> 4(0(4(1(3(2(x1)))))) 3(0(2(1(5(x1))))) -> 5(3(2(0(4(1(x1)))))) 3(0(4(0(2(x1))))) -> 0(3(4(0(4(2(x1)))))) 3(0(4(0(2(x1))))) -> 0(4(1(2(0(3(x1)))))) 3(0(5(1(4(x1))))) -> 3(0(4(1(1(5(x1)))))) 3(0(5(1(5(x1))))) -> 0(4(1(3(5(5(x1)))))) 3(2(4(1(2(x1))))) -> 3(1(2(2(5(4(x1)))))) 3(2(4(1(5(x1))))) -> 3(1(4(5(2(5(x1)))))) 3(4(0(1(2(x1))))) -> 0(4(2(0(3(1(x1)))))) 3(4(0(1(4(x1))))) -> 0(4(1(5(3(4(x1)))))) 3(4(0(1(5(x1))))) -> 0(4(1(5(5(3(x1)))))) 3(4(0(2(4(x1))))) -> 0(3(4(0(4(2(x1)))))) 3(4(1(2(4(x1))))) -> 0(4(1(2(4(3(x1)))))) 3(4(1(3(5(x1))))) -> 4(3(0(3(1(5(x1)))))) 3(4(3(0(2(x1))))) -> 3(3(0(4(1(2(x1)))))) 3(4(5(0(2(x1))))) -> 0(3(0(4(2(5(x1)))))) 3(5(0(2(2(x1))))) -> 0(3(2(5(2(5(x1)))))) 3(5(2(1(4(x1))))) -> 3(5(1(0(4(2(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(0(2(x1)))) -> 0^1(0(3(1(2(x1))))) 0^1(1(0(2(x1)))) -> 0^1(3(1(2(x1))))
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