theorem Th62:
  3 divides 2|^(2|^z) + (6*a-1)
  proof
A1: 3*(2*a) mod 3 = 0 by NAT_D:13;
A2: (6*a-1) mod 3 = ((6*a mod 3) - (1 mod 3)) mod 3 by INT_6:7
    .= (0 - 1) mod 3 by A1,NAT_D:24
    .= -1 mod 3
    .= 2 by Th9;
    ( 2|^(2|^z) + (6*a-1) ) mod 3
     = (2|^(2|^z) mod 3) + ((6*a-1) mod 3) mod 3 by NAT_D:66
    .= (1+2) mod 3 by A2,Th39
    .= 0 by NAT_D:25;
    hence thesis by INT_1:62;
  end;
