details

property	value
status	complete
benchmark	double.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n072.star.cs.uiowa.edu
space	AProVE_10

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	11.8854 seconds
cpu usage	36.4332
user time	34.9558
system time	1.47742
max virtual memory	3.7284032E7
max residence set size	3858136.0

stage attributes

key	value
starexec-result	NO

output

NO proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) AAECC Innermost [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) QReductionProof [EQUIVALENT, 0 ms] (11) QDP (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) MNOCProof [EQUIVALENT, 0 ms] (16) QDP (17) NonLoopProof [COMPLETE, 53 ms] (18) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(tt, x) -> f(isDouble(x), s(s(x))) isDouble(s(s(x))) -> isDouble(x) isDouble(0) -> tt Q is empty. ---------------------------------------- (1) AAECC Innermost (EQUIVALENT) We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is isDouble(s(s(x))) -> isDouble(x) isDouble(0) -> tt The TRS R 2 is f(tt, x) -> f(isDouble(x), s(s(x))) The signature Sigma is {f_2} ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(tt, x) -> f(isDouble(x), s(s(x))) isDouble(s(s(x))) -> isDouble(x) isDouble(0) -> tt The set Q consists of the following terms: f(tt, x0) isDouble(s(s(x0))) isDouble(0) ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: F(tt, x) -> F(isDouble(x), s(s(x))) F(tt, x) -> ISDOUBLE(x) ISDOUBLE(s(s(x))) -> ISDOUBLE(x) The TRS R consists of the following rules: f(tt, x) -> f(isDouble(x), s(s(x))) isDouble(s(s(x))) -> isDouble(x) isDouble(0) -> tt The set Q consists of the following terms: f(tt, x0) isDouble(s(s(x0))) isDouble(0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- popout

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actions

all output return to TRS Standard