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SRS_Standard 2019-03-29 03.29 pair #432288631
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
n143.star.cs.uiowa.edu
space
ICFP_2010
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
3.77461 seconds
cpu usage
11.4996
user time
10.8905
system time
0.60916
max virtual memory
1.9209704E7
max residence set size
1766348.0
stage attributes
key
value
starexec-result
YES
output
10.49/3.50 YES 11.23/3.68 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 11.23/3.68 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 11.23/3.68 11.23/3.68 11.23/3.68 Termination w.r.t. Q of the given QTRS could be proven: 11.23/3.68 11.23/3.68 (0) QTRS 11.23/3.68 (1) DependencyPairsProof [EQUIVALENT, 270 ms] 11.23/3.68 (2) QDP 11.23/3.68 (3) DependencyGraphProof [EQUIVALENT, 0 ms] 11.23/3.68 (4) AND 11.23/3.68 (5) QDP 11.23/3.68 (6) QDPSizeChangeProof [EQUIVALENT, 4 ms] 11.23/3.68 (7) YES 11.23/3.68 (8) QDP 11.23/3.68 (9) QDPSizeChangeProof [EQUIVALENT, 0 ms] 11.23/3.68 (10) YES 11.23/3.68 11.23/3.68 11.23/3.68 ---------------------------------------- 11.23/3.68 11.23/3.68 (0) 11.23/3.68 Obligation: 11.23/3.68 Q restricted rewrite system: 11.23/3.68 The TRS R consists of the following rules: 11.23/3.68 11.23/3.68 0(0(1(x1))) -> 0(2(3(0(1(x1))))) 11.23/3.68 0(0(1(x1))) -> 0(4(0(5(4(1(x1)))))) 11.23/3.68 0(0(1(x1))) -> 2(1(0(0(3(4(x1)))))) 11.23/3.68 0(0(1(x1))) -> 4(0(5(4(0(1(x1)))))) 11.23/3.68 0(1(0(x1))) -> 0(0(2(1(2(x1))))) 11.23/3.68 0(1(0(x1))) -> 1(0(0(5(4(x1))))) 11.23/3.68 0(1(0(x1))) -> 0(0(2(5(4(1(x1)))))) 11.23/3.68 0(1(1(x1))) -> 1(0(3(4(1(x1))))) 11.23/3.68 0(1(1(x1))) -> 5(0(3(4(1(1(x1)))))) 11.23/3.68 5(0(1(x1))) -> 0(5(4(1(x1)))) 11.23/3.68 5(0(1(x1))) -> 2(5(4(0(1(x1))))) 11.23/3.68 5(0(1(x1))) -> 5(0(2(1(2(x1))))) 11.23/3.68 5(0(1(x1))) -> 0(1(4(5(4(4(x1)))))) 11.23/3.68 5(0(1(x1))) -> 0(5(4(1(4(4(x1)))))) 11.23/3.68 5(0(1(x1))) -> 5(0(4(3(0(1(x1)))))) 11.23/3.68 5(1(0(x1))) -> 5(0(2(2(1(x1))))) 11.23/3.68 5(1(0(x1))) -> 5(0(5(4(1(x1))))) 11.23/3.68 5(1(0(x1))) -> 0(5(0(2(2(1(x1)))))) 11.23/3.68 5(1(0(x1))) -> 1(4(0(5(2(3(x1)))))) 11.23/3.68 5(1(0(x1))) -> 1(5(0(4(4(2(x1)))))) 11.23/3.68 5(1(0(x1))) -> 4(4(1(0(4(5(x1)))))) 11.23/3.68 5(1(1(x1))) -> 1(1(5(4(x1)))) 11.23/3.68 5(1(1(x1))) -> 5(4(1(1(x1)))) 11.23/3.68 5(1(1(x1))) -> 1(5(3(4(1(x1))))) 11.23/3.68 5(1(1(x1))) -> 1(1(4(5(4(4(x1)))))) 11.23/3.68 5(1(1(x1))) -> 3(5(2(3(1(1(x1)))))) 11.23/3.68 5(1(1(x1))) -> 4(1(2(1(5(4(x1)))))) 11.23/3.68 0(1(3(0(x1)))) -> 0(2(0(2(1(3(x1)))))) 11.23/3.68 0(1(5(0(x1)))) -> 0(0(5(4(1(5(x1)))))) 11.23/3.68 0(1(5(0(x1)))) -> 0(5(4(2(1(0(x1)))))) 11.23/3.68 0(3(0(1(x1)))) -> 0(0(4(1(3(0(x1)))))) 11.23/3.68 0(3(1(0(x1)))) -> 0(0(2(3(1(x1))))) 11.23/3.68 0(3(1(1(x1)))) -> 5(1(1(0(3(4(x1)))))) 11.23/3.68 5(0(1(0(x1)))) -> 5(0(0(4(1(3(x1)))))) 11.23/3.68 5(1(2(0(x1)))) -> 1(4(0(5(4(2(x1)))))) 11.23/3.68 5(1(2(0(x1)))) -> 5(0(4(2(2(1(x1)))))) 11.23/3.68 5(1(4(0(x1)))) -> 1(5(4(0(2(3(x1)))))) 11.23/3.68 5(1(4(0(x1)))) -> 4(5(2(1(3(0(x1)))))) 11.23/3.68 5(1(5(1(x1)))) -> 5(4(1(5(1(x1))))) 11.23/3.68 5(3(0(1(x1)))) -> 0(1(5(2(3(x1))))) 11.23/3.68 5(3(1(0(x1)))) -> 1(4(3(5(0(x1))))) 11.23/3.68 5(3(1(0(x1)))) -> 1(5(0(4(3(x1))))) 11.23/3.68 5(3(1(0(x1)))) -> 5(4(3(1(0(x1))))) 11.23/3.68 5(3(1(0(x1)))) -> 1(3(0(4(3(5(x1)))))) 11.23/3.68 5(3(1(1(x1)))) -> 1(1(5(3(3(4(x1)))))) 11.23/3.68 0(1(2(5(0(x1))))) -> 1(5(4(0(2(0(x1)))))) 11.23/3.68 0(1(4(2(0(x1))))) -> 1(0(4(2(3(0(x1)))))) 11.23/3.68 1(4(5(1(0(x1))))) -> 5(4(2(1(1(0(x1)))))) 11.23/3.68 5(0(1(4(0(x1))))) -> 1(4(5(4(0(0(x1)))))) 11.23/3.68 5(5(1(0(0(x1))))) -> 5(5(0(4(1(0(x1)))))) 11.23/3.68 11.23/3.68 Q is empty. 11.23/3.68 11.23/3.68 ---------------------------------------- 11.23/3.68 11.23/3.68 (1) DependencyPairsProof (EQUIVALENT) 11.23/3.68 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 11.23/3.68 ---------------------------------------- 11.23/3.68 11.23/3.68 (2) 11.23/3.68 Obligation: 11.23/3.68 Q DP problem: 11.23/3.68 The TRS P consists of the following rules: 11.23/3.68 11.23/3.68 0^1(0(1(x1))) -> 0^1(2(3(0(1(x1))))) 11.23/3.68 0^1(0(1(x1))) -> 0^1(4(0(5(4(1(x1)))))) 11.23/3.68 0^1(0(1(x1))) -> 0^1(5(4(1(x1)))) 11.23/3.68 0^1(0(1(x1))) -> 5^1(4(1(x1))) 11.23/3.68 0^1(0(1(x1))) -> 1^1(0(0(3(4(x1))))) 11.23/3.68 0^1(0(1(x1))) -> 0^1(0(3(4(x1)))) 11.23/3.68 0^1(0(1(x1))) -> 0^1(3(4(x1))) 11.23/3.68 0^1(0(1(x1))) -> 0^1(5(4(0(1(x1))))) 11.23/3.68 0^1(0(1(x1))) -> 5^1(4(0(1(x1))))
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