details

property	value
status	complete
benchmark	214091.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n159.star.cs.uiowa.edu
space	ICFP_2010

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	5.13924 seconds
cpu usage	16.8169
user time	16.1594
system time	0.657509
max virtual memory	2.0156868E7
max residence set size	2107980.0

stage attributes

key	value
starexec-result	YES

output

15.19/4.73 YES 15.56/4.76 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 15.56/4.76 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 15.56/4.76 15.56/4.76 15.56/4.76 Termination w.r.t. Q of the given QTRS could be proven: 15.56/4.76 15.56/4.76 (0) QTRS 15.56/4.76 (1) DependencyPairsProof [EQUIVALENT, 154 ms] 15.56/4.76 (2) QDP 15.56/4.76 (3) DependencyGraphProof [EQUIVALENT, 2 ms] 15.56/4.76 (4) AND 15.56/4.76 (5) QDP 15.56/4.76 (6) UsableRulesProof [EQUIVALENT, 0 ms] 15.56/4.76 (7) QDP 15.56/4.76 (8) QDPSizeChangeProof [EQUIVALENT, 0 ms] 15.56/4.76 (9) YES 15.56/4.76 (10) QDP 15.56/4.76 (11) UsableRulesProof [EQUIVALENT, 0 ms] 15.56/4.76 (12) QDP 15.56/4.76 (13) QDPSizeChangeProof [EQUIVALENT, 4 ms] 15.56/4.76 (14) YES 15.56/4.76 15.56/4.76 15.56/4.76 ---------------------------------------- 15.56/4.76 15.56/4.76 (0) 15.56/4.76 Obligation: 15.56/4.76 Q restricted rewrite system: 15.56/4.76 The TRS R consists of the following rules: 15.56/4.76 15.56/4.76 0(1(2(x1))) -> 0(2(1(0(x1)))) 15.56/4.76 0(1(2(x1))) -> 1(0(2(3(x1)))) 15.56/4.76 0(1(2(x1))) -> 0(2(4(1(5(x1))))) 15.56/4.76 0(1(2(x1))) -> 0(3(2(1(0(x1))))) 15.56/4.76 0(1(2(x1))) -> 1(0(3(2(3(x1))))) 15.56/4.76 0(1(2(x1))) -> 0(1(3(4(2(3(x1)))))) 15.56/4.76 0(5(2(x1))) -> 0(2(4(5(3(x1))))) 15.56/4.76 0(5(2(x1))) -> 5(4(2(3(0(4(x1)))))) 15.56/4.76 2(0(1(x1))) -> 3(0(2(1(x1)))) 15.56/4.76 2(0(1(x1))) -> 0(2(1(1(4(x1))))) 15.56/4.76 2(0(1(x1))) -> 0(3(2(4(1(x1))))) 15.56/4.76 2(0(1(x1))) -> 3(0(2(1(4(x1))))) 15.56/4.76 2(0(1(x1))) -> 0(2(2(3(4(1(x1)))))) 15.56/4.76 2(0(1(x1))) -> 0(3(2(3(1(1(x1)))))) 15.56/4.76 2(0(1(x1))) -> 4(0(4(2(1(4(x1)))))) 15.56/4.76 2(5(1(x1))) -> 0(2(1(5(1(x1))))) 15.56/4.76 2(5(1(x1))) -> 1(4(5(4(2(x1))))) 15.56/4.76 2(5(1(x1))) -> 5(0(2(1(4(x1))))) 15.56/4.76 2(5(1(x1))) -> 5(2(1(4(1(x1))))) 15.56/4.76 2(5(1(x1))) -> 1(5(0(2(4(1(x1)))))) 15.56/4.76 2(5(1(x1))) -> 5(2(1(1(1(1(x1)))))) 15.56/4.76 0(1(2(1(x1)))) -> 3(1(4(0(2(1(x1)))))) 15.56/4.76 0(1(3(1(x1)))) -> 5(0(3(1(1(x1))))) 15.56/4.76 0(1(3(1(x1)))) -> 1(0(3(4(2(1(x1)))))) 15.56/4.76 0(1(5(1(x1)))) -> 5(0(3(1(1(x1))))) 15.56/4.76 0(2(1(2(x1)))) -> 0(2(2(1(5(x1))))) 15.56/4.76 0(2(5(1(x1)))) -> 1(1(5(0(2(1(x1)))))) 15.56/4.76 0(5(3(1(x1)))) -> 0(1(4(4(3(5(x1)))))) 15.56/4.76 0(5(5(2(x1)))) -> 5(4(2(3(5(0(x1)))))) 15.56/4.76 2(0(1(2(x1)))) -> 0(2(3(2(1(1(x1)))))) 15.56/4.76 2(0(1(2(x1)))) -> 4(0(2(1(1(2(x1)))))) 15.56/4.76 2(0(4(1(x1)))) -> 3(0(2(4(1(x1))))) 15.56/4.76 2(0(5(1(x1)))) -> 5(4(2(1(0(x1))))) 15.56/4.76 2(2(5(1(x1)))) -> 3(2(2(4(5(1(x1)))))) 15.56/4.76 2(4(0(1(x1)))) -> 1(0(2(4(4(x1))))) 15.56/4.76 2(4(0(1(x1)))) -> 3(0(0(2(4(1(x1)))))) 15.56/4.76 2(4(0(1(x1)))) -> 5(4(0(2(1(1(x1)))))) 15.56/4.76 2(5(2(1(x1)))) -> 1(5(2(2(3(1(x1)))))) 15.56/4.76 2(5(4(1(x1)))) -> 4(5(2(1(4(4(x1)))))) 15.56/4.76 2(5(5(1(x1)))) -> 1(5(4(2(4(5(x1)))))) 15.56/4.76 2(5(5(2(x1)))) -> 5(5(2(3(2(x1))))) 15.56/4.76 0(1(3(0(1(x1))))) -> 0(3(1(0(1(1(x1)))))) 15.56/4.76 0(2(4(3(1(x1))))) -> 1(3(4(2(3(0(x1)))))) 15.56/4.76 0(2(4(3(1(x1))))) -> 4(0(3(2(1(0(x1)))))) 15.56/4.76 0(2(5(3(1(x1))))) -> 5(0(2(3(5(1(x1)))))) 15.56/4.76 2(0(5(4(1(x1))))) -> 0(4(5(3(2(1(x1)))))) 15.56/4.76 2(2(0(1(2(x1))))) -> 2(4(0(2(2(1(x1)))))) 15.56/4.76 2(4(0(5(1(x1))))) -> 1(4(5(0(4(2(x1)))))) 15.56/4.76 2(4(2(3(1(x1))))) -> 4(2(2(3(3(1(x1)))))) 15.56/4.76 2(5(2(0(1(x1))))) -> 0(2(4(1(5(2(x1)))))) 15.56/4.76 15.56/4.76 Q is empty. 15.56/4.76 15.56/4.76 ---------------------------------------- 15.56/4.76 15.56/4.76 (1) DependencyPairsProof (EQUIVALENT) 15.56/4.76 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 15.56/4.76 ---------------------------------------- 15.56/4.76 15.56/4.76 (2) 15.56/4.76 Obligation: 15.56/4.76 Q DP problem: 15.56/4.76 The TRS P consists of the following rules: 15.56/4.76 15.56/4.76 0^1(1(2(x1))) -> 0^1(2(1(0(x1)))) 15.56/4.76 0^1(1(2(x1))) -> 2^1(1(0(x1))) 15.56/4.76 0^1(1(2(x1))) -> 0^1(x1) 15.56/4.76 0^1(1(2(x1))) -> 0^1(2(3(x1))) 15.56/4.76 0^1(1(2(x1))) -> 2^1(3(x1)) popout

output may be truncated. 'popout' for the full output.

job log

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actions

all output return to SRS_Standard 2019-03-29 03.29