details

property	value
status	complete
benchmark	212693.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n169.star.cs.uiowa.edu
space	ICFP_2010

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	10.1692 seconds
cpu usage	20.5255
user time	19.8103
system time	0.715228
max virtual memory	3.7776952E7
max residence set size	2293004.0

stage attributes

key	value
starexec-result	YES

output

20.18/5.97 YES 20.18/6.03 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 20.18/6.03 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 20.18/6.03 20.18/6.03 20.18/6.03 Termination w.r.t. Q of the given QTRS could be proven: 20.18/6.03 20.18/6.03 (0) QTRS 20.18/6.03 (1) DependencyPairsProof [EQUIVALENT, 197 ms] 20.18/6.03 (2) QDP 20.18/6.03 (3) DependencyGraphProof [EQUIVALENT, 3 ms] 20.18/6.03 (4) AND 20.18/6.03 (5) QDP 20.18/6.03 (6) UsableRulesProof [EQUIVALENT, 0 ms] 20.18/6.03 (7) QDP 20.18/6.03 (8) QDPSizeChangeProof [EQUIVALENT, 0 ms] 20.18/6.03 (9) YES 20.18/6.03 (10) QDP 20.18/6.03 (11) UsableRulesProof [EQUIVALENT, 0 ms] 20.18/6.03 (12) QDP 20.18/6.03 (13) QDPOrderProof [EQUIVALENT, 15 ms] 20.18/6.03 (14) QDP 20.18/6.03 (15) PisEmptyProof [EQUIVALENT, 0 ms] 20.18/6.03 (16) YES 20.18/6.03 20.18/6.03 20.18/6.03 ---------------------------------------- 20.18/6.03 20.18/6.03 (0) 20.18/6.03 Obligation: 20.18/6.03 Q restricted rewrite system: 20.18/6.03 The TRS R consists of the following rules: 20.18/6.03 20.18/6.03 0(1(2(x1))) -> 0(0(2(1(x1)))) 20.18/6.03 0(1(2(x1))) -> 0(2(1(3(x1)))) 20.18/6.03 0(1(2(x1))) -> 0(0(2(1(4(4(x1)))))) 20.18/6.03 0(3(1(x1))) -> 0(1(3(4(0(x1))))) 20.18/6.03 0(3(1(x1))) -> 0(1(3(4(4(x1))))) 20.18/6.03 0(3(1(x1))) -> 1(3(4(4(4(0(x1)))))) 20.18/6.03 0(3(2(x1))) -> 0(2(1(3(x1)))) 20.18/6.03 0(3(2(x1))) -> 0(2(3(4(x1)))) 20.18/6.03 0(3(2(x1))) -> 0(0(2(4(3(x1))))) 20.18/6.03 0(3(2(x1))) -> 0(2(1(4(3(x1))))) 20.18/6.03 0(3(2(x1))) -> 0(2(4(3(3(x1))))) 20.18/6.03 0(3(2(x1))) -> 0(2(1(3(3(4(x1)))))) 20.18/6.03 0(3(2(x1))) -> 0(2(3(4(5(5(x1)))))) 20.18/6.03 0(3(2(x1))) -> 2(4(4(3(4(0(x1)))))) 20.18/6.03 0(4(1(x1))) -> 0(1(4(4(x1)))) 20.18/6.03 0(4(1(x1))) -> 0(2(1(4(x1)))) 20.18/6.03 0(4(2(x1))) -> 0(2(1(4(x1)))) 20.18/6.03 0(4(2(x1))) -> 0(2(3(4(x1)))) 20.18/6.03 0(4(2(x1))) -> 0(2(4(3(x1)))) 20.18/6.03 2(0(1(x1))) -> 5(0(2(1(x1)))) 20.18/6.03 2(3(1(x1))) -> 1(3(5(2(x1)))) 20.18/6.03 2(3(1(x1))) -> 0(2(1(3(5(x1))))) 20.18/6.03 2(3(1(x1))) -> 1(4(3(5(2(x1))))) 20.18/6.03 0(2(0(1(x1)))) -> 5(0(0(2(1(x1))))) 20.18/6.03 0(3(1(1(x1)))) -> 0(1(4(1(3(4(x1)))))) 20.18/6.03 0(3(2(1(x1)))) -> 0(0(3(4(2(1(x1)))))) 20.18/6.03 0(3(2(2(x1)))) -> 1(3(4(0(2(2(x1)))))) 20.18/6.03 0(4(1(2(x1)))) -> 1(4(0(2(5(x1))))) 20.18/6.03 0(4(3(2(x1)))) -> 2(3(4(4(0(0(x1)))))) 20.18/6.03 0(5(3(1(x1)))) -> 0(1(4(3(5(4(x1)))))) 20.18/6.03 0(5(3(1(x1)))) -> 0(1(5(3(4(0(x1)))))) 20.18/6.03 0(5(3(2(x1)))) -> 0(2(4(5(3(x1))))) 20.18/6.03 0(5(3(2(x1)))) -> 0(2(5(3(3(x1))))) 20.18/6.03 2(0(3(1(x1)))) -> 2(0(1(3(5(2(x1)))))) 20.18/6.03 2(0(4(1(x1)))) -> 2(0(1(4(5(x1))))) 20.18/6.03 2(5(3(2(x1)))) -> 2(5(2(3(3(x1))))) 20.18/6.03 2(5(4(2(x1)))) -> 0(2(5(2(4(x1))))) 20.18/6.03 0(0(3(2(1(x1))))) -> 0(0(1(3(5(2(x1)))))) 20.18/6.03 0(1(0(3(2(x1))))) -> 0(1(4(3(2(0(x1)))))) 20.18/6.03 0(1(0(3(2(x1))))) -> 2(3(1(0(0(5(x1)))))) 20.18/6.03 0(3(2(5(1(x1))))) -> 0(2(5(1(3(3(x1)))))) 20.18/6.03 0(5(1(1(2(x1))))) -> 0(2(4(1(1(5(x1)))))) 20.18/6.03 0(5(1(2(2(x1))))) -> 0(2(5(2(1(2(x1)))))) 20.18/6.03 0(5(3(2(1(x1))))) -> 0(1(3(4(2(5(x1)))))) 20.18/6.03 0(5(5(3(2(x1))))) -> 0(2(5(1(3(5(x1)))))) 20.18/6.03 2(0(3(1(1(x1))))) -> 2(1(0(1(3(4(x1)))))) 20.18/6.03 2(2(0(3(1(x1))))) -> 1(3(0(2(5(2(x1)))))) 20.18/6.03 2(2(0(5(1(x1))))) -> 2(0(2(1(5(1(x1)))))) 20.18/6.03 2(5(5(4(1(x1))))) -> 5(5(2(1(3(4(x1)))))) 20.18/6.03 20.18/6.03 Q is empty. 20.18/6.03 20.18/6.03 ---------------------------------------- 20.18/6.03 20.18/6.03 (1) DependencyPairsProof (EQUIVALENT) 20.18/6.03 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 20.18/6.03 ---------------------------------------- 20.18/6.03 20.18/6.03 (2) 20.18/6.03 Obligation: 20.18/6.03 Q DP problem: 20.18/6.03 The TRS P consists of the following rules: 20.18/6.03 20.18/6.03 0^1(1(2(x1))) -> 0^1(0(2(1(x1)))) 20.18/6.03 0^1(1(2(x1))) -> 0^1(2(1(x1))) 20.18/6.03 0^1(1(2(x1))) -> 2^1(1(x1)) 20.18/6.03 0^1(1(2(x1))) -> 0^1(2(1(3(x1)))) popout

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job log

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actions

all output return to SRS_Standard 2019-03-29 03.29