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	n093.star.cs.uiowa.edu
space	ICFP_2010

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	15.2136 seconds
cpu usage	56.1926
user time	54.099
system time	2.09364
max virtual memory	4.136578E7
max residence set size	4688352.0

stage attributes

key	value
starexec-result	YES

output

55.72/15.10 YES 55.98/15.13 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 55.98/15.13 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 55.98/15.13 55.98/15.13 55.98/15.13 Termination w.r.t. Q of the given QTRS could be proven: 55.98/15.13 55.98/15.13 (0) QTRS 55.98/15.13 (1) DependencyPairsProof [EQUIVALENT, 294 ms] 55.98/15.13 (2) QDP 55.98/15.13 (3) DependencyGraphProof [EQUIVALENT, 0 ms] 55.98/15.13 (4) QDP 55.98/15.13 (5) MRRProof [EQUIVALENT, 819 ms] 55.98/15.13 (6) QDP 55.98/15.13 (7) QDPOrderProof [EQUIVALENT, 59 ms] 55.98/15.13 (8) QDP 55.98/15.13 (9) QDPOrderProof [EQUIVALENT, 39 ms] 55.98/15.13 (10) QDP 55.98/15.13 (11) PisEmptyProof [EQUIVALENT, 0 ms] 55.98/15.13 (12) YES 55.98/15.13 55.98/15.13 55.98/15.13 ---------------------------------------- 55.98/15.13 55.98/15.13 (0) 55.98/15.13 Obligation: 55.98/15.13 Q restricted rewrite system: 55.98/15.13 The TRS R consists of the following rules: 55.98/15.13 55.98/15.13 0(1(2(2(x1)))) -> 0(1(0(2(2(x1))))) 55.98/15.13 0(1(2(2(x1)))) -> 0(1(2(3(2(x1))))) 55.98/15.13 0(1(2(2(x1)))) -> 0(2(2(1(3(x1))))) 55.98/15.13 0(1(2(2(x1)))) -> 1(0(3(2(2(x1))))) 55.98/15.13 0(1(2(2(x1)))) -> 1(2(0(3(2(x1))))) 55.98/15.13 0(1(2(2(x1)))) -> 1(3(0(2(2(x1))))) 55.98/15.13 0(1(2(2(x1)))) -> 1(3(2(0(2(x1))))) 55.98/15.13 0(1(2(2(x1)))) -> 0(1(0(4(2(2(x1)))))) 55.98/15.13 0(1(2(2(x1)))) -> 0(2(1(3(2(3(x1)))))) 55.98/15.13 0(1(2(2(x1)))) -> 1(2(1(0(4(2(x1)))))) 55.98/15.13 0(1(2(2(x1)))) -> 1(5(0(4(2(2(x1)))))) 55.98/15.13 0(1(2(2(x1)))) -> 2(0(3(1(3(2(x1)))))) 55.98/15.13 0(1(2(2(x1)))) -> 2(1(1(0(4(2(x1)))))) 55.98/15.13 0(1(2(2(x1)))) -> 2(1(3(0(2(0(x1)))))) 55.98/15.13 0(1(2(2(x1)))) -> 2(1(3(3(2(0(x1)))))) 55.98/15.13 0(1(2(2(x1)))) -> 2(1(5(3(0(2(x1)))))) 55.98/15.13 0(1(2(2(x1)))) -> 2(2(1(3(0(5(x1)))))) 55.98/15.13 0(1(2(2(x1)))) -> 2(4(1(3(2(0(x1)))))) 55.98/15.13 0(1(4(5(x1)))) -> 1(5(0(4(1(x1))))) 55.98/15.13 0(1(4(5(x1)))) -> 5(0(4(1(5(x1))))) 55.98/15.13 0(1(4(5(x1)))) -> 5(4(1(5(0(x1))))) 55.98/15.13 0(1(4(5(x1)))) -> 1(1(5(0(4(1(x1)))))) 55.98/15.13 0(1(4(5(x1)))) -> 5(4(1(5(5(0(x1)))))) 55.98/15.13 5(1(2(2(x1)))) -> 1(0(2(2(5(x1))))) 55.98/15.13 5(1(2(2(x1)))) -> 1(3(5(2(2(x1))))) 55.98/15.13 5(1(2(2(x1)))) -> 1(5(2(3(2(x1))))) 55.98/15.13 5(1(2(2(x1)))) -> 1(5(0(2(2(3(x1)))))) 55.98/15.13 5(1(2(2(x1)))) -> 2(1(0(3(2(5(x1)))))) 55.98/15.13 5(1(2(2(x1)))) -> 3(1(3(5(2(2(x1)))))) 55.98/15.13 5(1(2(2(x1)))) -> 4(1(3(2(2(5(x1)))))) 55.98/15.13 5(1(2(2(x1)))) -> 5(1(0(4(2(2(x1)))))) 55.98/15.13 5(1(2(2(x1)))) -> 5(1(2(0(4(2(x1)))))) 55.98/15.13 0(1(1(4(5(x1))))) -> 3(1(0(4(1(5(x1)))))) 55.98/15.13 0(1(2(2(2(x1))))) -> 1(0(2(2(5(2(x1)))))) 55.98/15.13 0(1(2(2(5(x1))))) -> 1(5(0(4(2(2(x1)))))) 55.98/15.13 0(1(2(4(5(x1))))) -> 2(5(1(0(4(5(x1)))))) 55.98/15.13 0(1(4(5(2(x1))))) -> 1(0(4(2(0(5(x1)))))) 55.98/15.13 0(1(4(5(5(x1))))) -> 5(0(4(0(1(5(x1)))))) 55.98/15.13 0(1(5(4(5(x1))))) -> 1(5(0(4(1(5(x1)))))) 55.98/15.13 0(5(1(2(2(x1))))) -> 0(1(3(2(5(2(x1)))))) 55.98/15.13 3(3(1(2(2(x1))))) -> 1(3(2(0(3(2(x1)))))) 55.98/15.13 3(4(4(0(5(x1))))) -> 3(5(4(5(0(4(x1)))))) 55.98/15.13 5(0(1(2(2(x1))))) -> 1(3(2(0(5(2(x1)))))) 55.98/15.13 5(1(2(2(5(x1))))) -> 1(5(2(3(2(5(x1)))))) 55.98/15.13 5(2(1(2(2(x1))))) -> 2(1(3(5(2(2(x1)))))) 55.98/15.13 5(2(4(0(5(x1))))) -> 0(4(2(5(5(5(x1)))))) 55.98/15.13 5(2(4(0(5(x1))))) -> 0(4(5(4(2(5(x1)))))) 55.98/15.13 55.98/15.13 Q is empty. 55.98/15.13 55.98/15.13 ---------------------------------------- 55.98/15.13 55.98/15.13 (1) DependencyPairsProof (EQUIVALENT) 55.98/15.13 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 55.98/15.13 ---------------------------------------- 55.98/15.13 55.98/15.13 (2) 55.98/15.13 Obligation: 55.98/15.13 Q DP problem: 55.98/15.13 The TRS P consists of the following rules: 55.98/15.13 55.98/15.13 0^1(1(2(2(x1)))) -> 0^1(1(0(2(2(x1))))) 55.98/15.13 0^1(1(2(2(x1)))) -> 0^1(2(2(x1))) 55.98/15.13 0^1(1(2(2(x1)))) -> 0^1(1(2(3(2(x1))))) 55.98/15.13 0^1(1(2(2(x1)))) -> 3^1(2(x1)) 55.98/15.13 0^1(1(2(2(x1)))) -> 0^1(2(2(1(3(x1))))) 55.98/15.13 0^1(1(2(2(x1)))) -> 3^1(x1) 55.98/15.13 0^1(1(2(2(x1)))) -> 0^1(3(2(2(x1)))) 55.98/15.13 0^1(1(2(2(x1)))) -> 3^1(2(2(x1))) 55.98/15.13 0^1(1(2(2(x1)))) -> 0^1(3(2(x1))) 55.98/15.13 0^1(1(2(2(x1)))) -> 3^1(0(2(2(x1)))) popout

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all output return to SRS_Standard 2019-03-29 03.29