details

property	value
status	complete
benchmark	3336.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n137.star.cs.uiowa.edu
space	ICFP_2010

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	5.02693 seconds
cpu usage	16.9222
user time	16.3744
system time	0.547756
max virtual memory	1.995662E7
max residence set size	1586912.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 170 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 185 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 2(5(x1)) -> 1(3(3(0(1(0(x1)))))) 2(5(x1)) -> 2(2(0(5(0(1(x1)))))) 3(5(x1)) -> 1(3(2(0(0(1(x1)))))) 3(5(x1)) -> 3(2(0(5(3(0(x1)))))) 4(5(x1)) -> 2(2(1(3(2(1(x1)))))) 4(5(x1)) -> 3(2(0(5(0(0(x1)))))) 1(2(5(x1))) -> 1(0(5(0(5(4(x1)))))) 1(2(5(x1))) -> 1(2(2(1(0(1(x1)))))) 1(2(5(x1))) -> 2(0(1(3(1(0(x1)))))) 1(4(5(x1))) -> 1(2(4(0(2(1(x1)))))) 2(5(1(x1))) -> 2(2(2(1(2(3(x1)))))) 2(5(2(x1))) -> 4(0(2(2(3(3(x1)))))) 2(5(3(x1))) -> 2(0(4(1(3(3(x1)))))) 2(5(4(x1))) -> 2(0(5(1(0(1(x1)))))) 3(2(5(x1))) -> 3(2(0(1(0(5(x1)))))) 3(4(2(x1))) -> 3(4(0(2(2(2(x1)))))) 3(5(1(x1))) -> 0(4(2(0(0(5(x1)))))) 3(5(1(x1))) -> 0(4(2(2(3(4(x1)))))) 3(5(1(x1))) -> 2(1(4(1(0(1(x1)))))) 3(5(2(x1))) -> 0(4(3(2(2(2(x1)))))) 3(5(2(x1))) -> 2(0(2(2(3(0(x1)))))) 3(5(2(x1))) -> 2(3(3(2(1(2(x1)))))) 3(5(3(x1))) -> 0(2(4(3(3(0(x1)))))) 3(5(3(x1))) -> 0(5(4(3(3(0(x1)))))) 3(5(3(x1))) -> 2(3(4(0(4(2(x1)))))) 3(5(4(x1))) -> 0(2(0(5(0(0(x1)))))) 3(5(4(x1))) -> 0(5(0(0(1(2(x1)))))) 3(5(5(x1))) -> 0(5(4(1(0(5(x1)))))) 4(5(1(x1))) -> 2(1(0(5(3(3(x1)))))) 4(5(2(x1))) -> 0(5(1(0(0(4(x1)))))) 4(5(4(x1))) -> 2(2(1(0(4(2(x1)))))) 4(5(4(x1))) -> 3(2(0(3(2(0(x1)))))) 5(5(3(x1))) -> 5(1(0(1(2(2(x1)))))) 5(5(4(x1))) -> 5(1(0(4(2(2(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: 2^1(5(x1)) -> 1^1(3(3(0(1(0(x1)))))) 2^1(5(x1)) -> 3^1(3(0(1(0(x1))))) 2^1(5(x1)) -> 3^1(0(1(0(x1)))) 2^1(5(x1)) -> 1^1(0(x1)) 2^1(5(x1)) -> 2^1(2(0(5(0(1(x1)))))) 2^1(5(x1)) -> 2^1(0(5(0(1(x1))))) 2^1(5(x1)) -> 5^1(0(1(x1))) 2^1(5(x1)) -> 1^1(x1) 3^1(5(x1)) -> 1^1(3(2(0(0(1(x1)))))) 3^1(5(x1)) -> 3^1(2(0(0(1(x1))))) 3^1(5(x1)) -> 2^1(0(0(1(x1)))) 3^1(5(x1)) -> 1^1(x1) 3^1(5(x1)) -> 3^1(2(0(5(3(0(x1)))))) 3^1(5(x1)) -> 2^1(0(5(3(0(x1))))) 3^1(5(x1)) -> 5^1(3(0(x1))) 3^1(5(x1)) -> 3^1(0(x1)) 4^1(5(x1)) -> 2^1(2(1(3(2(1(x1)))))) 4^1(5(x1)) -> 2^1(1(3(2(1(x1))))) 4^1(5(x1)) -> 1^1(3(2(1(x1)))) 4^1(5(x1)) -> 3^1(2(1(x1))) 4^1(5(x1)) -> 2^1(1(x1)) 4^1(5(x1)) -> 1^1(x1) 4^1(5(x1)) -> 3^1(2(0(5(0(0(x1)))))) 4^1(5(x1)) -> 2^1(0(5(0(0(x1))))) 4^1(5(x1)) -> 5^1(0(0(x1))) 1^1(2(5(x1))) -> 1^1(0(5(0(5(4(x1)))))) 1^1(2(5(x1))) -> 5^1(0(5(4(x1)))) popout

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