theorem Th18:
  len f > 1 & len g > 1 & Ant(Ant(f)) = Ant(Ant(g)) & 'not' Suc(
  Ant(f)) = Suc(Ant(g)) & Suc(f) = Suc(g) & Ant(f) |= Suc(f) & Ant(g) |= Suc(g)
  implies Ant(Ant(f)) |= Suc(f)
proof
  assume that
A1: len f > 1 and
A2: len g > 1 and
A3: Ant(Ant(f)) = Ant(Ant(g)) and
A4: 'not' Suc(Ant(f)) = Suc(Ant(g)) and
A5: Suc(f) = Suc(g) and
A6: Ant(f) |= Suc(f) and
A7: Ant(g) |= Suc(g);
  let A,J,v such that
A8: J,v |= Ant(Ant(f));
A9: len Ant(g) > 0 by A2,Th4;
A10: now
    assume not J,v |= Suc(Ant(f));
    then J,v |= Suc(Ant(g)) by A4,VALUAT_1:17;
    then J,v |= Ant(g) by A3,A9,A8,Th17;
    hence thesis by A5,A7;
  end;
A11: len Ant(f) > 0 by A1,Th4;
  now
    assume J,v |= Suc(Ant(f));
    then J,v |= Ant(f) by A11,A8,Th17;
    hence thesis by A6;
  end;
  hence thesis by A10;
end;
