details

property	value
status	complete
benchmark	n001.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n169.star.cs.uiowa.edu
space	Mixed_innermost

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.69035 seconds
cpu usage	6.85895
user time	6.51304
system time	0.345912
max virtual memory	1.914154E7
max residence set size	829600.0

stage attributes

key	value
starexec-result	NO

output

6.31/2.54 NO 6.64/2.65 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 6.64/2.65 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 6.64/2.65 6.64/2.65 6.64/2.65 Termination w.r.t. Q of the given QTRS could be disproven: 6.64/2.65 6.64/2.65 (0) QTRS 6.64/2.65 (1) DependencyPairsProof [EQUIVALENT, 0 ms] 6.64/2.65 (2) QDP 6.64/2.65 (3) DependencyGraphProof [EQUIVALENT, 0 ms] 6.64/2.65 (4) QDP 6.64/2.65 (5) UsableRulesProof [EQUIVALENT, 1 ms] 6.64/2.65 (6) QDP 6.64/2.65 (7) QReductionProof [EQUIVALENT, 0 ms] 6.64/2.65 (8) QDP 6.64/2.65 (9) NonTerminationLoopProof [COMPLETE, 0 ms] 6.64/2.65 (10) NO 6.64/2.65 6.64/2.65 6.64/2.65 ---------------------------------------- 6.64/2.65 6.64/2.65 (0) 6.64/2.65 Obligation: 6.64/2.65 Q restricted rewrite system: 6.64/2.65 The TRS R consists of the following rules: 6.64/2.65 6.64/2.65 h(f(f(x))) -> h(f(g(f(x)))) 6.64/2.65 f(g(f(x))) -> f(f(x)) 6.64/2.65 6.64/2.65 The set Q consists of the following terms: 6.64/2.65 6.64/2.65 h(f(f(x0))) 6.64/2.65 f(g(f(x0))) 6.64/2.65 6.64/2.65 6.64/2.65 ---------------------------------------- 6.64/2.65 6.64/2.65 (1) DependencyPairsProof (EQUIVALENT) 6.64/2.65 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 6.64/2.65 ---------------------------------------- 6.64/2.65 6.64/2.65 (2) 6.64/2.65 Obligation: 6.64/2.65 Q DP problem: 6.64/2.65 The TRS P consists of the following rules: 6.64/2.65 6.64/2.65 H(f(f(x))) -> H(f(g(f(x)))) 6.64/2.65 H(f(f(x))) -> F(g(f(x))) 6.64/2.65 F(g(f(x))) -> F(f(x)) 6.64/2.65 6.64/2.65 The TRS R consists of the following rules: 6.64/2.65 6.64/2.65 h(f(f(x))) -> h(f(g(f(x)))) 6.64/2.65 f(g(f(x))) -> f(f(x)) 6.64/2.65 6.64/2.65 The set Q consists of the following terms: 6.64/2.65 6.64/2.65 h(f(f(x0))) 6.64/2.65 f(g(f(x0))) 6.64/2.65 6.64/2.65 We have to consider all minimal (P,Q,R)-chains. 6.64/2.65 ---------------------------------------- 6.64/2.65 6.64/2.65 (3) DependencyGraphProof (EQUIVALENT) 6.64/2.65 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes. 6.64/2.65 ---------------------------------------- 6.64/2.65 6.64/2.65 (4) 6.64/2.65 Obligation: 6.64/2.65 Q DP problem: 6.64/2.65 The TRS P consists of the following rules: 6.64/2.65 6.64/2.65 H(f(f(x))) -> H(f(g(f(x)))) 6.64/2.65 6.64/2.65 The TRS R consists of the following rules: 6.64/2.65 6.64/2.65 h(f(f(x))) -> h(f(g(f(x)))) 6.64/2.65 f(g(f(x))) -> f(f(x)) 6.64/2.65 6.64/2.65 The set Q consists of the following terms: 6.64/2.65 6.64/2.65 h(f(f(x0))) 6.64/2.65 f(g(f(x0))) 6.64/2.65 6.64/2.65 We have to consider all minimal (P,Q,R)-chains. 6.64/2.65 ---------------------------------------- 6.64/2.65 6.64/2.65 (5) UsableRulesProof (EQUIVALENT) 6.64/2.65 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. 6.64/2.65 ---------------------------------------- 6.64/2.65 6.64/2.65 (6) 6.64/2.65 Obligation: 6.64/2.65 Q DP problem: 6.64/2.65 The TRS P consists of the following rules: 6.64/2.65 6.64/2.65 H(f(f(x))) -> H(f(g(f(x)))) 6.64/2.65 6.64/2.65 The TRS R consists of the following rules: popout

output may be truncated. 'popout' for the full output.

job log

popout

actions

all output return to TRS_Innermost 2019-03-28 22.12