details

property	value
status	complete
benchmark	size-12-alpha-3-num-191.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n047.star.cs.uiowa.edu
space	Waldmann_07_size12

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	8.40812 seconds
cpu usage	28.5827
user time	27.3071
system time	1.27556
max virtual memory	7.9164504E7
max residence set size	3565864.0

stage attributes

key	value
starexec-result	YES

output

26.96/7.77 YES 28.40/8.22 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 28.40/8.22 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 28.40/8.22 28.40/8.22 28.40/8.22 Termination w.r.t. Q of the given QTRS could be proven: 28.40/8.22 28.40/8.22 (0) QTRS 28.40/8.22 (1) QTRS Reverse [EQUIVALENT, 0 ms] 28.40/8.22 (2) QTRS 28.40/8.22 (3) DependencyPairsProof [EQUIVALENT, 30 ms] 28.40/8.22 (4) QDP 28.40/8.22 (5) DependencyGraphProof [EQUIVALENT, 1 ms] 28.40/8.22 (6) QDP 28.40/8.22 (7) QDPOrderProof [EQUIVALENT, 132 ms] 28.40/8.22 (8) QDP 28.40/8.22 (9) PisEmptyProof [EQUIVALENT, 0 ms] 28.40/8.22 (10) YES 28.40/8.22 28.40/8.22 28.40/8.22 ---------------------------------------- 28.40/8.22 28.40/8.22 (0) 28.40/8.22 Obligation: 28.40/8.22 Q restricted rewrite system: 28.40/8.22 The TRS R consists of the following rules: 28.40/8.22 28.40/8.22 a(x1) -> x1 28.40/8.22 a(b(x1)) -> b(b(a(a(c(x1))))) 28.40/8.22 b(x1) -> x1 28.40/8.22 c(b(c(x1))) -> x1 28.40/8.22 28.40/8.22 Q is empty. 28.40/8.22 28.40/8.22 ---------------------------------------- 28.40/8.22 28.40/8.22 (1) QTRS Reverse (EQUIVALENT) 28.40/8.22 We applied the QTRS Reverse Processor [REVERSE]. 28.40/8.22 ---------------------------------------- 28.40/8.22 28.40/8.22 (2) 28.40/8.22 Obligation: 28.40/8.22 Q restricted rewrite system: 28.40/8.22 The TRS R consists of the following rules: 28.40/8.22 28.40/8.22 a(x1) -> x1 28.40/8.22 b(a(x1)) -> c(a(a(b(b(x1))))) 28.40/8.22 b(x1) -> x1 28.40/8.22 c(b(c(x1))) -> x1 28.40/8.22 28.40/8.22 Q is empty. 28.40/8.22 28.40/8.22 ---------------------------------------- 28.40/8.22 28.40/8.22 (3) DependencyPairsProof (EQUIVALENT) 28.40/8.22 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 28.40/8.22 ---------------------------------------- 28.40/8.22 28.40/8.22 (4) 28.40/8.22 Obligation: 28.40/8.22 Q DP problem: 28.40/8.22 The TRS P consists of the following rules: 28.40/8.22 28.40/8.22 B(a(x1)) -> C(a(a(b(b(x1))))) 28.40/8.22 B(a(x1)) -> A(a(b(b(x1)))) 28.40/8.22 B(a(x1)) -> A(b(b(x1))) 28.40/8.22 B(a(x1)) -> B(b(x1)) 28.40/8.22 B(a(x1)) -> B(x1) 28.40/8.22 28.40/8.22 The TRS R consists of the following rules: 28.40/8.22 28.40/8.22 a(x1) -> x1 28.40/8.22 b(a(x1)) -> c(a(a(b(b(x1))))) 28.40/8.22 b(x1) -> x1 28.40/8.22 c(b(c(x1))) -> x1 28.40/8.22 28.40/8.22 Q is empty. 28.40/8.22 We have to consider all minimal (P,Q,R)-chains. 28.40/8.22 ---------------------------------------- 28.40/8.22 28.40/8.22 (5) DependencyGraphProof (EQUIVALENT) 28.40/8.22 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes. 28.40/8.22 ---------------------------------------- 28.40/8.22 28.40/8.22 (6) 28.40/8.22 Obligation: 28.40/8.22 Q DP problem: 28.40/8.22 The TRS P consists of the following rules: 28.40/8.22 28.40/8.22 B(a(x1)) -> B(x1) 28.40/8.22 B(a(x1)) -> B(b(x1)) 28.40/8.22 28.40/8.22 The TRS R consists of the following rules: 28.40/8.22 28.40/8.22 a(x1) -> x1 28.40/8.22 b(a(x1)) -> c(a(a(b(b(x1))))) 28.40/8.22 b(x1) -> x1 28.40/8.22 c(b(c(x1))) -> x1 28.40/8.22 28.40/8.22 Q is empty. popout

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all output return to SRS_Standard 2019-03-29 03.29