details

property	value
status	complete
benchmark	test9.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n139.star.cs.uiowa.edu
space	Mixed_innermost

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.91296 seconds
cpu usage	4.0769
user time	3.87264
system time	0.204259
max virtual memory	1.8417272E7
max residence set size	295692.0

stage attributes

key	value
starexec-result	YES

output

3.95/1.88 YES 3.95/1.89 proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml 3.95/1.89 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 3.95/1.89 3.95/1.89 3.95/1.89 Termination w.r.t. Q of the given QTRS could be proven: 3.95/1.89 3.95/1.89 (0) QTRS 3.95/1.89 (1) DependencyPairsProof [EQUIVALENT, 0 ms] 3.95/1.89 (2) QDP 3.95/1.89 (3) UsableRulesProof [EQUIVALENT, 0 ms] 3.95/1.89 (4) QDP 3.95/1.89 (5) QReductionProof [EQUIVALENT, 0 ms] 3.95/1.89 (6) QDP 3.95/1.89 (7) TransformationProof [EQUIVALENT, 0 ms] 3.95/1.89 (8) QDP 3.95/1.89 (9) DependencyGraphProof [EQUIVALENT, 0 ms] 3.95/1.89 (10) TRUE 3.95/1.89 3.95/1.89 3.95/1.89 ---------------------------------------- 3.95/1.89 3.95/1.89 (0) 3.95/1.89 Obligation: 3.95/1.89 Q restricted rewrite system: 3.95/1.89 The TRS R consists of the following rules: 3.95/1.89 3.95/1.89 f(0, 1, X) -> h(X, X) 3.95/1.89 h(0, X) -> f(0, X, X) 3.95/1.89 g(X, Y) -> X 3.95/1.89 g(X, Y) -> Y 3.95/1.89 3.95/1.89 The set Q consists of the following terms: 3.95/1.89 3.95/1.89 f(0, 1, x0) 3.95/1.89 h(0, x0) 3.95/1.89 g(x0, x1) 3.95/1.89 3.95/1.89 3.95/1.89 ---------------------------------------- 3.95/1.89 3.95/1.89 (1) DependencyPairsProof (EQUIVALENT) 3.95/1.89 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 3.95/1.89 ---------------------------------------- 3.95/1.89 3.95/1.89 (2) 3.95/1.89 Obligation: 3.95/1.89 Q DP problem: 3.95/1.89 The TRS P consists of the following rules: 3.95/1.89 3.95/1.89 F(0, 1, X) -> H(X, X) 3.95/1.89 H(0, X) -> F(0, X, X) 3.95/1.89 3.95/1.89 The TRS R consists of the following rules: 3.95/1.89 3.95/1.89 f(0, 1, X) -> h(X, X) 3.95/1.89 h(0, X) -> f(0, X, X) 3.95/1.89 g(X, Y) -> X 3.95/1.89 g(X, Y) -> Y 3.95/1.89 3.95/1.89 The set Q consists of the following terms: 3.95/1.89 3.95/1.89 f(0, 1, x0) 3.95/1.89 h(0, x0) 3.95/1.89 g(x0, x1) 3.95/1.89 3.95/1.89 We have to consider all minimal (P,Q,R)-chains. 3.95/1.89 ---------------------------------------- 3.95/1.89 3.95/1.89 (3) UsableRulesProof (EQUIVALENT) 3.95/1.89 As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. 3.95/1.89 ---------------------------------------- 3.95/1.89 3.95/1.89 (4) 3.95/1.89 Obligation: 3.95/1.89 Q DP problem: 3.95/1.89 The TRS P consists of the following rules: 3.95/1.89 3.95/1.89 F(0, 1, X) -> H(X, X) 3.95/1.89 H(0, X) -> F(0, X, X) 3.95/1.89 3.95/1.89 R is empty. 3.95/1.89 The set Q consists of the following terms: 3.95/1.89 3.95/1.89 f(0, 1, x0) 3.95/1.89 h(0, x0) 3.95/1.89 g(x0, x1) 3.95/1.89 3.95/1.89 We have to consider all minimal (P,Q,R)-chains. 3.95/1.89 ---------------------------------------- 3.95/1.89 3.95/1.89 (5) QReductionProof (EQUIVALENT) 3.95/1.89 We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 3.95/1.89 3.95/1.89 f(0, 1, x0) 3.95/1.89 h(0, x0) 3.95/1.89 g(x0, x1) 3.95/1.89 3.95/1.89 3.95/1.89 ---------------------------------------- popout

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job log

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actions

all output return to TRS_Innermost 2019-03-28 22.12