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SRS_Standard 2019-03-29 03.29 pair #432288547
details
property
value
status
complete
benchmark
211915.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n084.star.cs.uiowa.edu
space
ICFP_2010
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
14.6381 seconds
cpu usage
52.3578
user time
50.4621
system time
1.8957
max virtual memory
8.155324E7
max residence set size
4420476.0
stage attributes
key
value
starexec-result
YES
output
51.79/14.45 YES 52.03/14.50 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 52.03/14.50 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 52.03/14.50 52.03/14.50 52.03/14.50 Termination w.r.t. Q of the given QTRS could be proven: 52.03/14.50 52.03/14.50 (0) QTRS 52.03/14.50 (1) DependencyPairsProof [EQUIVALENT, 280 ms] 52.03/14.50 (2) QDP 52.03/14.50 (3) DependencyGraphProof [EQUIVALENT, 4 ms] 52.03/14.50 (4) AND 52.03/14.50 (5) QDP 52.03/14.50 (6) UsableRulesProof [EQUIVALENT, 0 ms] 52.03/14.50 (7) QDP 52.03/14.50 (8) MRRProof [EQUIVALENT, 0 ms] 52.03/14.50 (9) QDP 52.03/14.50 (10) QDPOrderProof [EQUIVALENT, 2 ms] 52.03/14.50 (11) QDP 52.03/14.50 (12) PisEmptyProof [EQUIVALENT, 0 ms] 52.03/14.50 (13) YES 52.03/14.50 (14) QDP 52.03/14.50 (15) QDPOrderProof [EQUIVALENT, 59 ms] 52.03/14.50 (16) QDP 52.03/14.50 (17) QDPOrderProof [EQUIVALENT, 91 ms] 52.03/14.50 (18) QDP 52.03/14.50 (19) QDPOrderProof [EQUIVALENT, 61 ms] 52.03/14.50 (20) QDP 52.03/14.50 (21) PisEmptyProof [EQUIVALENT, 0 ms] 52.03/14.50 (22) YES 52.03/14.50 52.03/14.50 52.03/14.50 ---------------------------------------- 52.03/14.50 52.03/14.50 (0) 52.03/14.50 Obligation: 52.03/14.50 Q restricted rewrite system: 52.03/14.50 The TRS R consists of the following rules: 52.03/14.50 52.03/14.50 0(1(2(x1))) -> 1(3(0(2(x1)))) 52.03/14.50 0(1(2(x1))) -> 3(1(2(0(x1)))) 52.03/14.50 0(1(2(x1))) -> 2(0(4(1(3(x1))))) 52.03/14.50 0(1(2(x1))) -> 3(0(2(1(3(x1))))) 52.03/14.50 0(1(2(x1))) -> 3(3(0(2(1(x1))))) 52.03/14.50 0(1(2(x1))) -> 1(3(3(0(2(3(x1)))))) 52.03/14.50 0(1(2(x1))) -> 2(0(4(3(1(3(x1)))))) 52.03/14.50 0(1(2(x1))) -> 3(0(1(3(1(2(x1)))))) 52.03/14.50 0(1(2(x1))) -> 3(0(5(3(1(2(x1)))))) 52.03/14.50 0(5(2(x1))) -> 1(5(0(2(x1)))) 52.03/14.50 0(5(2(x1))) -> 3(5(0(2(x1)))) 52.03/14.50 0(5(2(x1))) -> 3(0(5(1(2(x1))))) 52.03/14.50 0(5(2(x1))) -> 3(5(3(0(2(x1))))) 52.03/14.50 0(5(2(x1))) -> 1(5(5(3(0(2(x1)))))) 52.03/14.50 0(5(2(x1))) -> 5(0(4(3(1(2(x1)))))) 52.03/14.50 1(5(2(x1))) -> 5(3(1(2(x1)))) 52.03/14.50 1(5(2(x1))) -> 1(5(3(1(2(x1))))) 52.03/14.50 0(0(5(4(x1)))) -> 5(3(0(4(0(0(x1)))))) 52.03/14.50 0(1(0(1(x1)))) -> 1(1(3(0(3(0(x1)))))) 52.03/14.50 0(1(1(2(x1)))) -> 2(1(1(3(0(2(x1)))))) 52.03/14.50 0(1(2(5(x1)))) -> 3(0(5(1(2(x1))))) 52.03/14.50 0(1(2(5(x1)))) -> 3(5(3(1(2(0(x1)))))) 52.03/14.50 0(1(3(2(x1)))) -> 3(0(2(1(3(5(x1)))))) 52.03/14.50 0(1(4(2(x1)))) -> 3(0(0(4(1(2(x1)))))) 52.03/14.50 0(1(4(2(x1)))) -> 4(0(1(3(1(2(x1)))))) 52.03/14.50 0(5(2(4(x1)))) -> 5(5(3(0(2(4(x1)))))) 52.03/14.50 0(5(2(5(x1)))) -> 0(2(3(5(5(x1))))) 52.03/14.50 0(5(3(2(x1)))) -> 3(0(2(1(5(x1))))) 52.03/14.50 0(5(3(2(x1)))) -> 5(3(1(4(0(2(x1)))))) 52.03/14.50 1(0(3(4(x1)))) -> 1(0(4(1(3(x1))))) 52.03/14.50 1(5(3(2(x1)))) -> 3(1(3(5(2(x1))))) 52.03/14.50 5(0(1(2(x1)))) -> 1(5(0(4(1(2(x1)))))) 52.03/14.50 5(1(0(2(x1)))) -> 3(1(5(0(2(x1))))) 52.03/14.50 5(1(1(2(x1)))) -> 1(3(5(2(1(3(x1)))))) 52.03/14.50 5(1(5(2(x1)))) -> 1(3(5(5(2(x1))))) 52.03/14.50 0(0(1(3(2(x1))))) -> 0(3(1(2(0(3(x1)))))) 52.03/14.50 0(0(5(3(2(x1))))) -> 0(3(0(3(2(5(x1)))))) 52.03/14.50 0(1(0(3(4(x1))))) -> 4(0(1(3(0(3(x1)))))) 52.03/14.50 0(1(0(5(4(x1))))) -> 3(0(4(0(1(5(x1)))))) 52.03/14.50 0(1(2(4(4(x1))))) -> 0(4(1(2(1(4(x1)))))) 52.03/14.50 0(1(5(3(2(x1))))) -> 1(2(3(0(4(5(x1)))))) 52.03/14.50 0(5(4(2(4(x1))))) -> 4(5(3(0(4(2(x1)))))) 52.03/14.50 0(5(4(4(2(x1))))) -> 0(4(4(3(5(2(x1)))))) 52.03/14.50 1(0(1(3(4(x1))))) -> 4(1(1(3(0(3(x1)))))) 52.03/14.50 1(0(3(4(5(x1))))) -> 2(1(3(0(4(5(x1)))))) 52.03/14.50 1(1(3(4(2(x1))))) -> 2(5(1(3(1(4(x1)))))) 52.03/14.50 1(5(3(2(2(x1))))) -> 2(3(5(3(1(2(x1)))))) 52.03/14.50 5(0(5(2(4(x1))))) -> 5(1(5(0(4(2(x1)))))) 52.03/14.50 5(1(0(5(2(x1))))) -> 1(5(0(2(3(5(x1)))))) 52.03/14.50 5(1(1(2(1(x1))))) -> 3(1(2(5(1(1(x1)))))) 52.03/14.50 52.03/14.50 Q is empty. 52.03/14.50 52.03/14.50 ---------------------------------------- 52.03/14.50 52.03/14.50 (1) DependencyPairsProof (EQUIVALENT) 52.03/14.50 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 52.03/14.50 ---------------------------------------- 52.03/14.50 52.03/14.50 (2) 52.03/14.50 Obligation:
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