details

property	value
status	complete
benchmark	213218.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n140.star.cs.uiowa.edu
space	ICFP_2010

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	3.26257 seconds
cpu usage	9.96206
user time	9.52317
system time	0.438896
max virtual memory	1.9614948E7
max residence set size	1315180.0

stage attributes

key	value
starexec-result	YES

output

YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 203 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 1 ms] (7) QDP (8) QDPSizeChangeProof [EQUIVALENT, 0 ms] (9) YES (10) QDP (11) UsableRulesProof [EQUIVALENT, 0 ms] (12) QDP (13) QDPSizeChangeProof [EQUIVALENT, 0 ms] (14) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(1(2(x1))) -> 0(2(1(1(x1)))) 2(3(1(x1))) -> 3(4(2(1(x1)))) 0(1(2(0(x1)))) -> 0(0(2(1(1(x1))))) 0(1(2(0(x1)))) -> 0(2(1(1(0(x1))))) 0(1(2(1(x1)))) -> 0(2(1(1(1(x1))))) 0(1(2(4(x1)))) -> 4(0(2(1(1(x1))))) 0(1(2(5(x1)))) -> 0(2(5(1(1(x1))))) 0(1(2(5(x1)))) -> 0(4(2(1(5(x1))))) 0(1(3(1(x1)))) -> 0(0(3(1(1(x1))))) 0(1(3(1(x1)))) -> 3(0(2(1(1(1(x1)))))) 0(1(4(1(x1)))) -> 4(0(1(1(1(x1))))) 0(1(4(5(x1)))) -> 0(4(1(1(5(x1))))) 0(2(0(1(x1)))) -> 0(0(2(4(1(x1))))) 0(5(3(1(x1)))) -> 5(0(3(1(1(1(x1)))))) 0(5(3(2(x1)))) -> 0(3(4(2(5(x1))))) 2(3(2(0(x1)))) -> 2(2(1(3(0(x1))))) 2(4(5(2(x1)))) -> 2(1(4(2(5(x1))))) 3(0(1(2(x1)))) -> 3(4(0(2(1(x1))))) 4(4(3(2(x1)))) -> 4(3(4(2(1(x1))))) 4(5(3(1(x1)))) -> 3(5(4(4(1(x1))))) 4(5(3(1(x1)))) -> 4(3(1(5(1(x1))))) 4(5(3(2(x1)))) -> 3(4(2(5(4(x1))))) 4(5(3(2(x1)))) -> 3(5(4(2(1(x1))))) 0(1(0(2(2(x1))))) -> 0(0(2(1(4(2(x1)))))) 0(1(0(3(1(x1))))) -> 0(3(4(0(1(1(x1)))))) 0(1(4(5(1(x1))))) -> 4(0(2(5(1(1(x1)))))) 0(1(5(0(1(x1))))) -> 0(0(1(5(5(1(x1)))))) 0(2(3(1(3(x1))))) -> 0(3(4(2(1(3(x1)))))) 0(4(1(5(2(x1))))) -> 0(4(2(5(1(1(x1)))))) 0(5(1(3(2(x1))))) -> 3(0(2(1(1(5(x1)))))) 0(5(2(0(4(x1))))) -> 0(4(2(5(0(4(x1)))))) 0(5(3(1(5(x1))))) -> 5(0(3(4(1(5(x1)))))) 0(5(3(2(5(x1))))) -> 0(4(3(5(2(5(x1)))))) 2(0(1(3(1(x1))))) -> 1(1(4(3(0(2(x1)))))) 2(0(1(3(5(x1))))) -> 2(5(1(1(0(3(x1)))))) 2(0(1(5(3(x1))))) -> 2(1(1(3(0(5(x1)))))) 2(3(1(0(1(x1))))) -> 0(3(2(1(1(5(x1)))))) 2(3(1(4(1(x1))))) -> 3(4(2(5(1(1(x1)))))) 2(3(5(1(2(x1))))) -> 2(3(2(1(5(1(x1)))))) 3(0(1(2(0(x1))))) -> 3(4(0(2(1(0(x1)))))) 3(2(0(1(0(x1))))) -> 3(2(1(5(0(0(x1)))))) 3(2(3(5(1(x1))))) -> 3(3(4(2(1(5(x1)))))) 3(5(3(1(3(x1))))) -> 3(5(4(3(1(3(x1)))))) 4(0(3(3(1(x1))))) -> 0(3(4(3(1(1(x1)))))) 4(4(3(2(5(x1))))) -> 3(4(2(4(1(5(x1)))))) 4(5(0(5(2(x1))))) -> 0(4(2(1(5(5(x1)))))) 4(5(2(3(1(x1))))) -> 5(3(4(3(2(1(x1)))))) 4(5(2(5(2(x1))))) -> 4(2(5(5(2(1(x1)))))) 4(5(3(5(1(x1))))) -> 4(5(5(4(3(1(x1)))))) Q is empty. ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: 0^1(1(2(x1))) -> 0^1(2(1(1(x1)))) 0^1(1(2(x1))) -> 2^1(1(1(x1))) 2^1(3(1(x1))) -> 3^1(4(2(1(x1)))) 2^1(3(1(x1))) -> 4^1(2(1(x1))) 2^1(3(1(x1))) -> 2^1(1(x1)) 0^1(1(2(0(x1)))) -> 0^1(0(2(1(1(x1))))) popout

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

job log

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actions

all output return to SRS Standard