details

property	value
status	complete
benchmark	01.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n160.star.cs.uiowa.edu
space	Mixed_SRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	12.2186 seconds
cpu usage	38.299
user time	36.7932
system time	1.50581
max virtual memory	5.9799748E7
max residence set size	4169812.0

stage attributes

key	value
starexec-result	YES

output

37.33/10.45 YES 38.14/10.64 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 38.14/10.64 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 38.14/10.64 38.14/10.64 38.14/10.64 Termination w.r.t. Q of the given QTRS could be proven: 38.14/10.64 38.14/10.64 (0) QTRS 38.14/10.64 (1) QTRS Reverse [EQUIVALENT, 0 ms] 38.14/10.64 (2) QTRS 38.14/10.64 (3) DependencyPairsProof [EQUIVALENT, 16 ms] 38.14/10.64 (4) QDP 38.14/10.64 (5) MRRProof [EQUIVALENT, 71 ms] 38.14/10.64 (6) QDP 38.14/10.64 (7) QDPOrderProof [EQUIVALENT, 43 ms] 38.14/10.64 (8) QDP 38.14/10.64 (9) DependencyGraphProof [EQUIVALENT, 0 ms] 38.14/10.64 (10) TRUE 38.14/10.64 38.14/10.64 38.14/10.64 ---------------------------------------- 38.14/10.64 38.14/10.64 (0) 38.14/10.64 Obligation: 38.14/10.64 Q restricted rewrite system: 38.14/10.64 The TRS R consists of the following rules: 38.14/10.64 38.14/10.64 a(a(a(a(x1)))) -> a(b(a(b(x1)))) 38.14/10.64 b(a(b(x1))) -> a(b(a(x1))) 38.14/10.64 38.14/10.64 Q is empty. 38.14/10.64 38.14/10.64 ---------------------------------------- 38.14/10.64 38.14/10.64 (1) QTRS Reverse (EQUIVALENT) 38.14/10.64 We applied the QTRS Reverse Processor [REVERSE]. 38.14/10.64 ---------------------------------------- 38.14/10.64 38.14/10.64 (2) 38.14/10.64 Obligation: 38.14/10.64 Q restricted rewrite system: 38.14/10.64 The TRS R consists of the following rules: 38.14/10.64 38.14/10.64 a(a(a(a(x1)))) -> b(a(b(a(x1)))) 38.14/10.64 b(a(b(x1))) -> a(b(a(x1))) 38.14/10.64 38.14/10.64 Q is empty. 38.14/10.64 38.14/10.64 ---------------------------------------- 38.14/10.64 38.14/10.64 (3) DependencyPairsProof (EQUIVALENT) 38.14/10.64 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 38.14/10.64 ---------------------------------------- 38.14/10.64 38.14/10.64 (4) 38.14/10.64 Obligation: 38.14/10.64 Q DP problem: 38.14/10.64 The TRS P consists of the following rules: 38.14/10.64 38.14/10.64 A(a(a(a(x1)))) -> B(a(b(a(x1)))) 38.14/10.64 A(a(a(a(x1)))) -> A(b(a(x1))) 38.14/10.64 A(a(a(a(x1)))) -> B(a(x1)) 38.14/10.64 B(a(b(x1))) -> A(b(a(x1))) 38.14/10.64 B(a(b(x1))) -> B(a(x1)) 38.14/10.64 B(a(b(x1))) -> A(x1) 38.14/10.64 38.14/10.64 The TRS R consists of the following rules: 38.14/10.64 38.14/10.64 a(a(a(a(x1)))) -> b(a(b(a(x1)))) 38.14/10.64 b(a(b(x1))) -> a(b(a(x1))) 38.14/10.64 38.14/10.64 Q is empty. 38.14/10.64 We have to consider all minimal (P,Q,R)-chains. 38.14/10.64 ---------------------------------------- 38.14/10.64 38.14/10.64 (5) MRRProof (EQUIVALENT) 38.14/10.64 By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. 38.14/10.64 38.14/10.64 Strictly oriented dependency pairs: 38.14/10.64 38.14/10.64 A(a(a(a(x1)))) -> A(b(a(x1))) 38.14/10.64 A(a(a(a(x1)))) -> B(a(x1)) 38.14/10.64 B(a(b(x1))) -> B(a(x1)) 38.14/10.64 B(a(b(x1))) -> A(x1) 38.14/10.64 38.14/10.64 38.14/10.64 Used ordering: Polynomial interpretation [POLO]: 38.14/10.64 38.14/10.64 POL(A(x_1)) = 2 + 2*x_1 38.14/10.64 POL(B(x_1)) = 2 + 2*x_1 38.14/10.64 POL(a(x_1)) = 1 + x_1 38.14/10.64 POL(b(x_1)) = 1 + x_1 38.14/10.64 38.14/10.64 38.14/10.64 ---------------------------------------- 38.14/10.64 38.14/10.64 (6) 38.14/10.64 Obligation: 38.14/10.64 Q DP problem: 38.14/10.64 The TRS P consists of the following rules: popout

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

job log

popout

actions

all output return to SRS_Standard 2019-03-29 03.29