reserve c,c1,c2,x,y,z,z1,z2 for set;
reserve C1,C2,C3 for non empty set;

theorem
  for f,g be RMembership_Func of C1,C2 holds converse (f \+\ g) = (
  converse f) \+\ (converse g)
proof
  let f,g be RMembership_Func of C1,C2;
  converse (f \+\ g) = max(converse min(f,1_minus g),converse min(1_minus
  f,g)) by Th7
    .= max(min(converse f,converse 1_minus g), converse min(1_minus f,g)) by
Th8
    .= max(min(converse f,converse 1_minus g), min(converse 1_minus f,
  converse g)) by Th8
    .= max(min(converse f,1_minus (converse g)), min(converse 1_minus f,
  converse g)) by Th6
    .= max(min(converse f,1_minus (converse g)), min(1_minus converse f,
  converse g)) by Th6;
  hence thesis;
end;
