details

property	value
status	complete
benchmark	size-12-alpha-3-num-399.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n119.star.cs.uiowa.edu
space	Waldmann_07_size12

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	6.3879 seconds
cpu usage	21.3578
user time	20.3915
system time	0.966294
max virtual memory	8.2081064E7
max residence set size	2809304.0

stage attributes

key	value
starexec-result	YES

output

21.01/6.26 YES 21.01/6.30 proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml 21.01/6.30 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 21.01/6.30 21.01/6.30 21.01/6.30 Termination w.r.t. Q of the given QTRS could be proven: 21.01/6.30 21.01/6.30 (0) QTRS 21.01/6.30 (1) QTRS Reverse [EQUIVALENT, 0 ms] 21.01/6.30 (2) QTRS 21.01/6.30 (3) DependencyPairsProof [EQUIVALENT, 5 ms] 21.01/6.30 (4) QDP 21.01/6.30 (5) QDPOrderProof [EQUIVALENT, 65 ms] 21.01/6.30 (6) QDP 21.01/6.30 (7) QDPOrderProof [EQUIVALENT, 84 ms] 21.01/6.30 (8) QDP 21.01/6.30 (9) DependencyGraphProof [EQUIVALENT, 0 ms] 21.01/6.30 (10) TRUE 21.01/6.30 21.01/6.30 21.01/6.30 ---------------------------------------- 21.01/6.30 21.01/6.30 (0) 21.01/6.30 Obligation: 21.01/6.30 Q restricted rewrite system: 21.01/6.30 The TRS R consists of the following rules: 21.01/6.30 21.01/6.30 a(x1) -> b(x1) 21.01/6.30 b(b(x1)) -> c(x1) 21.01/6.30 c(a(c(x1))) -> a(b(c(a(x1)))) 21.01/6.30 21.01/6.30 Q is empty. 21.01/6.30 21.01/6.30 ---------------------------------------- 21.01/6.30 21.01/6.30 (1) QTRS Reverse (EQUIVALENT) 21.01/6.30 We applied the QTRS Reverse Processor [REVERSE]. 21.01/6.30 ---------------------------------------- 21.01/6.30 21.01/6.30 (2) 21.01/6.30 Obligation: 21.01/6.30 Q restricted rewrite system: 21.01/6.30 The TRS R consists of the following rules: 21.01/6.30 21.01/6.30 a(x1) -> b(x1) 21.01/6.30 b(b(x1)) -> c(x1) 21.01/6.30 c(a(c(x1))) -> a(c(b(a(x1)))) 21.01/6.30 21.01/6.30 Q is empty. 21.01/6.30 21.01/6.30 ---------------------------------------- 21.01/6.30 21.01/6.30 (3) DependencyPairsProof (EQUIVALENT) 21.01/6.30 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 21.01/6.30 ---------------------------------------- 21.01/6.30 21.01/6.30 (4) 21.01/6.30 Obligation: 21.01/6.30 Q DP problem: 21.01/6.30 The TRS P consists of the following rules: 21.01/6.30 21.01/6.30 A(x1) -> B(x1) 21.01/6.30 B(b(x1)) -> C(x1) 21.01/6.30 C(a(c(x1))) -> A(c(b(a(x1)))) 21.01/6.30 C(a(c(x1))) -> C(b(a(x1))) 21.01/6.30 C(a(c(x1))) -> B(a(x1)) 21.01/6.30 C(a(c(x1))) -> A(x1) 21.01/6.30 21.01/6.30 The TRS R consists of the following rules: 21.01/6.30 21.01/6.30 a(x1) -> b(x1) 21.01/6.30 b(b(x1)) -> c(x1) 21.01/6.30 c(a(c(x1))) -> a(c(b(a(x1)))) 21.01/6.30 21.01/6.30 Q is empty. 21.01/6.30 We have to consider all minimal (P,Q,R)-chains. 21.01/6.30 ---------------------------------------- 21.01/6.30 21.01/6.30 (5) QDPOrderProof (EQUIVALENT) 21.01/6.30 We use the reduction pair processor [LPAR04,JAR06]. 21.01/6.30 21.01/6.30 21.01/6.30 The following pairs can be oriented strictly and are deleted. 21.01/6.30 21.01/6.30 C(a(c(x1))) -> C(b(a(x1))) 21.01/6.30 C(a(c(x1))) -> B(a(x1)) 21.01/6.30 C(a(c(x1))) -> A(x1) 21.01/6.30 The remaining pairs can at least be oriented weakly. 21.01/6.30 Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation: 21.01/6.30 21.01/6.30 POL( A_1(x_1) ) = 2x_1 21.01/6.30 POL( B_1(x_1) ) = 2x_1 21.01/6.30 POL( C_1(x_1) ) = 2x_1 + 2 21.01/6.30 POL( b_1(x_1) ) = x_1 + 1 21.01/6.30 POL( c_1(x_1) ) = x_1 + 2 21.01/6.30 POL( a_1(x_1) ) = x_1 + 1 21.01/6.30 21.01/6.30 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 21.01/6.30 21.01/6.30 b(b(x1)) -> c(x1) popout

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all output return to SRS_Standard 2019-03-29 03.29