details

property	value
status	complete
benchmark	#4.26.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n114.star.cs.uiowa.edu
space	AG01_innermost

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.23414 seconds
cpu usage	4.3659
user time	4.14054
system time	0.225366
max virtual memory	1.8414384E7
max residence set size	311008.0

stage attributes

key	value
starexec-result	YES

output

4.23/2.18 YES 4.23/2.19 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 4.23/2.19 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 4.23/2.19 4.23/2.19 4.23/2.19 Termination w.r.t. Q of the given QTRS could be proven: 4.23/2.19 4.23/2.19 (0) QTRS 4.23/2.19 (1) DependencyPairsProof [EQUIVALENT, 20 ms] 4.23/2.19 (2) QDP 4.23/2.19 (3) DependencyGraphProof [EQUIVALENT, 0 ms] 4.23/2.19 (4) AND 4.23/2.19 (5) QDP 4.23/2.19 (6) UsableRulesProof [EQUIVALENT, 0 ms] 4.23/2.19 (7) QDP 4.23/2.19 (8) QReductionProof [EQUIVALENT, 0 ms] 4.23/2.19 (9) QDP 4.23/2.19 (10) QDPSizeChangeProof [EQUIVALENT, 0 ms] 4.23/2.19 (11) YES 4.23/2.19 (12) QDP 4.23/2.19 (13) UsableRulesProof [EQUIVALENT, 0 ms] 4.23/2.19 (14) QDP 4.23/2.19 (15) QReductionProof [EQUIVALENT, 0 ms] 4.23/2.19 (16) QDP 4.23/2.19 (17) QDPOrderProof [EQUIVALENT, 13 ms] 4.23/2.19 (18) QDP 4.23/2.19 (19) DependencyGraphProof [EQUIVALENT, 0 ms] 4.23/2.19 (20) TRUE 4.23/2.19 4.23/2.19 4.23/2.19 ---------------------------------------- 4.23/2.19 4.23/2.19 (0) 4.23/2.19 Obligation: 4.23/2.19 Q restricted rewrite system: 4.23/2.19 The TRS R consists of the following rules: 4.23/2.19 4.23/2.19 p(0) -> 0 4.23/2.19 p(s(x)) -> x 4.23/2.19 le(0, y) -> true 4.23/2.19 le(s(x), 0) -> false 4.23/2.19 le(s(x), s(y)) -> le(x, y) 4.23/2.19 minus(x, y) -> if(le(x, y), x, y) 4.23/2.19 if(true, x, y) -> 0 4.23/2.19 if(false, x, y) -> s(minus(p(x), y)) 4.23/2.19 4.23/2.19 The set Q consists of the following terms: 4.23/2.19 4.23/2.19 p(0) 4.23/2.19 p(s(x0)) 4.23/2.19 le(0, x0) 4.23/2.19 le(s(x0), 0) 4.23/2.19 le(s(x0), s(x1)) 4.23/2.19 minus(x0, x1) 4.23/2.19 if(true, x0, x1) 4.23/2.19 if(false, x0, x1) 4.23/2.19 4.23/2.19 4.23/2.19 ---------------------------------------- 4.23/2.19 4.23/2.19 (1) DependencyPairsProof (EQUIVALENT) 4.23/2.19 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 4.23/2.19 ---------------------------------------- 4.23/2.19 4.23/2.19 (2) 4.23/2.19 Obligation: 4.23/2.19 Q DP problem: 4.23/2.19 The TRS P consists of the following rules: 4.23/2.19 4.23/2.19 LE(s(x), s(y)) -> LE(x, y) 4.23/2.19 MINUS(x, y) -> IF(le(x, y), x, y) 4.23/2.19 MINUS(x, y) -> LE(x, y) 4.23/2.19 IF(false, x, y) -> MINUS(p(x), y) 4.23/2.19 IF(false, x, y) -> P(x) 4.23/2.19 4.23/2.19 The TRS R consists of the following rules: 4.23/2.19 4.23/2.19 p(0) -> 0 4.23/2.19 p(s(x)) -> x 4.23/2.19 le(0, y) -> true 4.23/2.19 le(s(x), 0) -> false 4.23/2.19 le(s(x), s(y)) -> le(x, y) 4.23/2.19 minus(x, y) -> if(le(x, y), x, y) 4.23/2.19 if(true, x, y) -> 0 4.23/2.19 if(false, x, y) -> s(minus(p(x), y)) 4.23/2.19 4.23/2.19 The set Q consists of the following terms: 4.23/2.19 4.23/2.19 p(0) 4.23/2.19 p(s(x0)) 4.23/2.19 le(0, x0) 4.23/2.19 le(s(x0), 0) 4.23/2.19 le(s(x0), s(x1)) 4.23/2.19 minus(x0, x1) 4.23/2.19 if(true, x0, x1) 4.23/2.19 if(false, x0, x1) 4.23/2.19 4.23/2.19 We have to consider all minimal (P,Q,R)-chains. 4.23/2.19 ---------------------------------------- 4.23/2.19 popout

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all output return to TRS_Innermost 2019-03-28 22.12