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	n141.star.cs.uiowa.edu
space	ICFP_2010

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	13.5458 seconds
cpu usage	49.3668
user time	47.3215
system time	2.04536
max virtual memory	4.010046E7
max residence set size	4649608.0

stage attributes

key	value
starexec-result	YES

output

YES proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 164 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 6 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 290 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 340 ms] (10) QDP (11) PisEmptyProof [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(1(2(2(x1)))) -> 0(1(0(2(2(x1))))) 0(1(2(2(x1)))) -> 0(1(2(3(2(x1))))) 0(1(2(2(x1)))) -> 0(2(2(1(3(x1))))) 0(1(2(2(x1)))) -> 1(0(3(2(2(x1))))) 0(1(2(2(x1)))) -> 1(2(0(3(2(x1))))) 0(1(2(2(x1)))) -> 1(3(0(2(2(x1))))) 0(1(2(2(x1)))) -> 1(3(2(0(2(x1))))) 0(1(2(2(x1)))) -> 0(1(0(4(2(2(x1)))))) 0(1(2(2(x1)))) -> 0(2(1(3(2(3(x1)))))) 0(1(2(2(x1)))) -> 1(2(1(0(4(2(x1)))))) 0(1(2(2(x1)))) -> 1(5(0(4(2(2(x1)))))) 0(1(2(2(x1)))) -> 2(0(3(1(3(2(x1)))))) 0(1(2(2(x1)))) -> 2(1(1(0(4(2(x1)))))) 0(1(2(2(x1)))) -> 2(1(3(0(2(0(x1)))))) 0(1(2(2(x1)))) -> 2(1(3(3(2(0(x1)))))) 0(1(2(2(x1)))) -> 2(1(5(3(0(2(x1)))))) 0(1(2(2(x1)))) -> 2(2(1(3(0(5(x1)))))) 0(1(2(2(x1)))) -> 2(4(1(3(2(0(x1)))))) 0(1(4(5(x1)))) -> 1(5(0(4(1(x1))))) 0(1(4(5(x1)))) -> 5(0(4(1(5(x1))))) 0(1(4(5(x1)))) -> 5(4(1(5(0(x1))))) 0(1(4(5(x1)))) -> 1(1(5(0(4(1(x1)))))) 0(1(4(5(x1)))) -> 5(4(1(5(5(0(x1)))))) 5(1(2(2(x1)))) -> 1(0(2(2(5(x1))))) 5(1(2(2(x1)))) -> 1(3(5(2(2(x1))))) 5(1(2(2(x1)))) -> 1(5(2(3(2(x1))))) 5(1(2(2(x1)))) -> 1(5(0(2(2(3(x1)))))) 5(1(2(2(x1)))) -> 2(1(0(3(2(5(x1)))))) 5(1(2(2(x1)))) -> 3(1(3(5(2(2(x1)))))) 5(1(2(2(x1)))) -> 4(1(3(2(2(5(x1)))))) 5(1(2(2(x1)))) -> 5(1(0(4(2(2(x1)))))) 5(1(2(2(x1)))) -> 5(1(2(0(4(2(x1)))))) 0(1(1(4(5(x1))))) -> 3(1(0(4(1(5(x1)))))) 0(1(2(2(2(x1))))) -> 1(0(2(2(5(2(x1)))))) 0(1(2(2(5(x1))))) -> 1(5(0(4(2(2(x1)))))) 0(1(2(4(5(x1))))) -> 2(5(1(0(4(5(x1)))))) 0(1(4(5(2(x1))))) -> 1(0(4(2(0(5(x1)))))) 0(1(4(5(5(x1))))) -> 5(0(4(0(1(5(x1)))))) 0(1(5(4(5(x1))))) -> 1(5(0(4(1(5(x1)))))) 0(5(1(2(2(x1))))) -> 0(1(3(2(5(2(x1)))))) 3(3(1(2(2(x1))))) -> 1(3(2(0(3(2(x1)))))) 3(4(4(0(5(x1))))) -> 3(5(4(5(0(4(x1)))))) 5(0(1(2(2(x1))))) -> 1(3(2(0(5(2(x1)))))) 5(1(2(2(5(x1))))) -> 1(5(2(3(2(5(x1)))))) 5(2(1(2(2(x1))))) -> 2(1(3(5(2(2(x1)))))) 5(2(4(0(5(x1))))) -> 0(4(2(5(5(5(x1)))))) 5(2(4(0(5(x1))))) -> 0(4(5(4(2(5(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(2(x1)))) -> 0^1(1(0(2(2(x1))))) 0^1(1(2(2(x1)))) -> 0^1(2(2(x1))) 0^1(1(2(2(x1)))) -> 0^1(1(2(3(2(x1))))) 0^1(1(2(2(x1)))) -> 3^1(2(x1)) 0^1(1(2(2(x1)))) -> 0^1(2(2(1(3(x1))))) 0^1(1(2(2(x1)))) -> 3^1(x1) 0^1(1(2(2(x1)))) -> 0^1(3(2(2(x1)))) 0^1(1(2(2(x1)))) -> 3^1(2(2(x1))) 0^1(1(2(2(x1)))) -> 0^1(3(2(x1))) 0^1(1(2(2(x1)))) -> 3^1(0(2(2(x1)))) popout

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

job log

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actions

all output return to SRS Standard