details

property	value
status	complete
benchmark	3336.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n081.star.cs.uiowa.edu
space	ICFP_2010

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	7.97351 seconds
cpu usage	27.9067
user time	27.0792
system time	0.827517
max virtual memory	3.6842976E7
max residence set size	2471608.0

stage attributes

key	value
starexec-result	YES

output

27.31/7.75 YES 27.66/7.83 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 27.66/7.83 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 27.66/7.83 27.66/7.83 27.66/7.83 Termination w.r.t. Q of the given QTRS could be proven: 27.66/7.83 27.66/7.83 (0) QTRS 27.66/7.83 (1) DependencyPairsProof [EQUIVALENT, 184 ms] 27.66/7.83 (2) QDP 27.66/7.83 (3) DependencyGraphProof [EQUIVALENT, 0 ms] 27.66/7.83 (4) QDP 27.66/7.83 (5) QDPOrderProof [EQUIVALENT, 255 ms] 27.66/7.83 (6) QDP 27.66/7.83 (7) DependencyGraphProof [EQUIVALENT, 0 ms] 27.66/7.83 (8) TRUE 27.66/7.83 27.66/7.83 27.66/7.83 ---------------------------------------- 27.66/7.83 27.66/7.83 (0) 27.66/7.83 Obligation: 27.66/7.83 Q restricted rewrite system: 27.66/7.83 The TRS R consists of the following rules: 27.66/7.83 27.66/7.83 2(5(x1)) -> 1(3(3(0(1(0(x1)))))) 27.66/7.83 2(5(x1)) -> 2(2(0(5(0(1(x1)))))) 27.66/7.83 3(5(x1)) -> 1(3(2(0(0(1(x1)))))) 27.66/7.83 3(5(x1)) -> 3(2(0(5(3(0(x1)))))) 27.66/7.83 4(5(x1)) -> 2(2(1(3(2(1(x1)))))) 27.66/7.83 4(5(x1)) -> 3(2(0(5(0(0(x1)))))) 27.66/7.83 1(2(5(x1))) -> 1(0(5(0(5(4(x1)))))) 27.66/7.83 1(2(5(x1))) -> 1(2(2(1(0(1(x1)))))) 27.66/7.83 1(2(5(x1))) -> 2(0(1(3(1(0(x1)))))) 27.66/7.83 1(4(5(x1))) -> 1(2(4(0(2(1(x1)))))) 27.66/7.83 2(5(1(x1))) -> 2(2(2(1(2(3(x1)))))) 27.66/7.83 2(5(2(x1))) -> 4(0(2(2(3(3(x1)))))) 27.66/7.83 2(5(3(x1))) -> 2(0(4(1(3(3(x1)))))) 27.66/7.83 2(5(4(x1))) -> 2(0(5(1(0(1(x1)))))) 27.66/7.83 3(2(5(x1))) -> 3(2(0(1(0(5(x1)))))) 27.66/7.83 3(4(2(x1))) -> 3(4(0(2(2(2(x1)))))) 27.66/7.83 3(5(1(x1))) -> 0(4(2(0(0(5(x1)))))) 27.66/7.83 3(5(1(x1))) -> 0(4(2(2(3(4(x1)))))) 27.66/7.83 3(5(1(x1))) -> 2(1(4(1(0(1(x1)))))) 27.66/7.83 3(5(2(x1))) -> 0(4(3(2(2(2(x1)))))) 27.66/7.83 3(5(2(x1))) -> 2(0(2(2(3(0(x1)))))) 27.66/7.83 3(5(2(x1))) -> 2(3(3(2(1(2(x1)))))) 27.66/7.83 3(5(3(x1))) -> 0(2(4(3(3(0(x1)))))) 27.66/7.83 3(5(3(x1))) -> 0(5(4(3(3(0(x1)))))) 27.66/7.83 3(5(3(x1))) -> 2(3(4(0(4(2(x1)))))) 27.66/7.83 3(5(4(x1))) -> 0(2(0(5(0(0(x1)))))) 27.66/7.83 3(5(4(x1))) -> 0(5(0(0(1(2(x1)))))) 27.66/7.83 3(5(5(x1))) -> 0(5(4(1(0(5(x1)))))) 27.66/7.83 4(5(1(x1))) -> 2(1(0(5(3(3(x1)))))) 27.66/7.83 4(5(2(x1))) -> 0(5(1(0(0(4(x1)))))) 27.66/7.83 4(5(4(x1))) -> 2(2(1(0(4(2(x1)))))) 27.66/7.83 4(5(4(x1))) -> 3(2(0(3(2(0(x1)))))) 27.66/7.83 5(5(3(x1))) -> 5(1(0(1(2(2(x1)))))) 27.66/7.83 5(5(4(x1))) -> 5(1(0(4(2(2(x1)))))) 27.66/7.83 27.66/7.83 Q is empty. 27.66/7.83 27.66/7.83 ---------------------------------------- 27.66/7.83 27.66/7.83 (1) DependencyPairsProof (EQUIVALENT) 27.66/7.83 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 27.66/7.83 ---------------------------------------- 27.66/7.83 27.66/7.83 (2) 27.66/7.83 Obligation: 27.66/7.83 Q DP problem: 27.66/7.83 The TRS P consists of the following rules: 27.66/7.83 27.66/7.83 2^1(5(x1)) -> 1^1(3(3(0(1(0(x1)))))) 27.66/7.83 2^1(5(x1)) -> 3^1(3(0(1(0(x1))))) 27.66/7.83 2^1(5(x1)) -> 3^1(0(1(0(x1)))) 27.66/7.83 2^1(5(x1)) -> 1^1(0(x1)) 27.66/7.83 2^1(5(x1)) -> 2^1(2(0(5(0(1(x1)))))) 27.66/7.83 2^1(5(x1)) -> 2^1(0(5(0(1(x1))))) 27.66/7.83 2^1(5(x1)) -> 5^1(0(1(x1))) 27.66/7.83 2^1(5(x1)) -> 1^1(x1) 27.66/7.83 3^1(5(x1)) -> 1^1(3(2(0(0(1(x1)))))) 27.66/7.83 3^1(5(x1)) -> 3^1(2(0(0(1(x1))))) 27.66/7.83 3^1(5(x1)) -> 2^1(0(0(1(x1)))) 27.66/7.83 3^1(5(x1)) -> 1^1(x1) 27.66/7.83 3^1(5(x1)) -> 3^1(2(0(5(3(0(x1)))))) 27.66/7.83 3^1(5(x1)) -> 2^1(0(5(3(0(x1))))) 27.66/7.83 3^1(5(x1)) -> 5^1(3(0(x1))) 27.66/7.83 3^1(5(x1)) -> 3^1(0(x1)) 27.66/7.83 4^1(5(x1)) -> 2^1(2(1(3(2(1(x1)))))) 27.66/7.83 4^1(5(x1)) -> 2^1(1(3(2(1(x1))))) 27.66/7.83 4^1(5(x1)) -> 1^1(3(2(1(x1)))) 27.66/7.83 4^1(5(x1)) -> 3^1(2(1(x1))) 27.66/7.83 4^1(5(x1)) -> 2^1(1(x1)) 27.66/7.83 4^1(5(x1)) -> 1^1(x1) 27.66/7.83 4^1(5(x1)) -> 3^1(2(0(5(0(0(x1)))))) 27.66/7.83 4^1(5(x1)) -> 2^1(0(5(0(0(x1))))) 27.66/7.83 4^1(5(x1)) -> 5^1(0(0(x1))) 27.66/7.83 1^1(2(5(x1))) -> 1^1(0(5(0(5(4(x1)))))) 27.66/7.83 1^1(2(5(x1))) -> 5^1(0(5(4(x1)))) popout

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all output return to SRS_Standard 2019-03-29 03.29