let nn, nn' be Element of NAT ; :: thesis: ( nn = (2 * nn') + 1 & nn' > 0 implies 6 + ((6 * ([\(log 2,nn')/] + 1)) + 1) = (6 * ([\(log 2,nn)/] + 1)) + 1 )
assume A1: ( nn = (2 * nn') + 1 & nn' > 0 ) ; :: thesis: 6 + ((6 * ([\(log 2,nn')/] + 1)) + 1) = (6 * ([\(log 2,nn)/] + 1)) + 1
set F = [\(log 2,nn')/];
thus 6 + ((6 * ([\(log 2,nn')/] + 1)) + 1) = (6 * (1 + ([\(log 2,nn')/] + 1))) + 1
.= (6 * ([\(log 2,nn)/] + 1)) + 1 by A1, PRE_FF:16 ; :: thesis: verum