details

property	value
status	complete
benchmark	size-11-alpha-3-num-9.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n032.star.cs.uiowa.edu
space	Waldmann_07_size11

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	9.41676 seconds
cpu usage	24.5643
user time	23.3655
system time	1.19878
max virtual memory	9.9914476E7
max residence set size	3148108.0

stage attributes

key	value
starexec-result	YES

output

22.17/6.53 YES 24.03/7.15 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 24.03/7.15 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 24.03/7.15 24.03/7.15 24.03/7.15 Termination w.r.t. Q of the given QTRS could be proven: 24.03/7.15 24.03/7.15 (0) QTRS 24.03/7.15 (1) QTRS Reverse [EQUIVALENT, 0 ms] 24.03/7.15 (2) QTRS 24.03/7.15 (3) DependencyPairsProof [EQUIVALENT, 2 ms] 24.03/7.15 (4) QDP 24.03/7.15 (5) DependencyGraphProof [EQUIVALENT, 1 ms] 24.03/7.15 (6) QDP 24.03/7.15 (7) QDPOrderProof [EQUIVALENT, 92 ms] 24.03/7.15 (8) QDP 24.03/7.15 (9) UsableRulesProof [EQUIVALENT, 0 ms] 24.03/7.15 (10) QDP 24.03/7.15 (11) QDPSizeChangeProof [EQUIVALENT, 0 ms] 24.03/7.15 (12) YES 24.03/7.15 24.03/7.15 24.03/7.15 ---------------------------------------- 24.03/7.15 24.03/7.15 (0) 24.03/7.15 Obligation: 24.03/7.15 Q restricted rewrite system: 24.03/7.15 The TRS R consists of the following rules: 24.03/7.15 24.03/7.15 a(x1) -> b(x1) 24.03/7.15 a(c(x1)) -> c(c(a(a(b(x1))))) 24.03/7.15 b(b(x1)) -> x1 24.03/7.15 24.03/7.15 Q is empty. 24.03/7.15 24.03/7.15 ---------------------------------------- 24.03/7.15 24.03/7.15 (1) QTRS Reverse (EQUIVALENT) 24.03/7.15 We applied the QTRS Reverse Processor [REVERSE]. 24.03/7.15 ---------------------------------------- 24.03/7.15 24.03/7.15 (2) 24.03/7.15 Obligation: 24.03/7.15 Q restricted rewrite system: 24.03/7.15 The TRS R consists of the following rules: 24.03/7.15 24.03/7.15 a(x1) -> b(x1) 24.03/7.15 c(a(x1)) -> b(a(a(c(c(x1))))) 24.03/7.15 b(b(x1)) -> x1 24.03/7.15 24.03/7.15 Q is empty. 24.03/7.15 24.03/7.15 ---------------------------------------- 24.03/7.15 24.03/7.15 (3) DependencyPairsProof (EQUIVALENT) 24.03/7.15 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 24.03/7.15 ---------------------------------------- 24.03/7.15 24.03/7.15 (4) 24.03/7.15 Obligation: 24.03/7.15 Q DP problem: 24.03/7.15 The TRS P consists of the following rules: 24.03/7.15 24.03/7.15 A(x1) -> B(x1) 24.03/7.15 C(a(x1)) -> B(a(a(c(c(x1))))) 24.03/7.15 C(a(x1)) -> A(a(c(c(x1)))) 24.03/7.15 C(a(x1)) -> A(c(c(x1))) 24.03/7.15 C(a(x1)) -> C(c(x1)) 24.03/7.15 C(a(x1)) -> C(x1) 24.03/7.15 24.03/7.15 The TRS R consists of the following rules: 24.03/7.15 24.03/7.15 a(x1) -> b(x1) 24.03/7.15 c(a(x1)) -> b(a(a(c(c(x1))))) 24.03/7.15 b(b(x1)) -> x1 24.03/7.15 24.03/7.15 Q is empty. 24.03/7.15 We have to consider all minimal (P,Q,R)-chains. 24.03/7.15 ---------------------------------------- 24.03/7.15 24.03/7.15 (5) DependencyGraphProof (EQUIVALENT) 24.03/7.15 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes. 24.03/7.15 ---------------------------------------- 24.03/7.15 24.03/7.15 (6) 24.03/7.15 Obligation: 24.03/7.15 Q DP problem: 24.03/7.15 The TRS P consists of the following rules: 24.03/7.15 24.03/7.15 C(a(x1)) -> C(x1) 24.03/7.15 C(a(x1)) -> C(c(x1)) 24.03/7.15 24.03/7.15 The TRS R consists of the following rules: 24.03/7.15 24.03/7.15 a(x1) -> b(x1) 24.03/7.15 c(a(x1)) -> b(a(a(c(c(x1))))) 24.03/7.15 b(b(x1)) -> x1 24.03/7.15 24.03/7.15 Q is empty. 24.03/7.15 We have to consider all minimal (P,Q,R)-chains. popout

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