let nn9 be Element of NAT ; ( nn9 > 0 implies ( [\(log 2,nn9)/] is Element of NAT & (6 * ([\(log 2,nn9)/] + 1)) + 1 > 0 ) )
assume
nn9 > 0
; ( [\(log 2,nn9)/] is Element of NAT & (6 * ([\(log 2,nn9)/] + 1)) + 1 > 0 )
then A1:
nn9 >= 0 + 1
by NAT_1:13;
log 2,1 = 0
by POWER:59;
then
log 2,nn9 >= 0
by A1, PRE_FF:12;
then
[\(log 2,nn9)/] >= [\0 /]
by PRE_FF:11;
then
[\(log 2,nn9)/] >= 0
by INT_1:47;
then reconsider F = [\(log 2,nn9)/] as Element of NAT by INT_1:16;
F is Element of NAT
;
hence
[\(log 2,nn9)/] is Element of NAT
; (6 * ([\(log 2,nn9)/] + 1)) + 1 > 0
(6 * F) + (6 * 1) >= 0
by NAT_1:12;
hence
(6 * ([\(log 2,nn9)/] + 1)) + 1 > 0
by NAT_1:13; verum