theorem Th50:
  S = C1 \/ C2 implies E-bound S = max(E-bound C1, E-bound C2)
proof
  assume
A1: S = C1 \/ C2;
A2: E-bound C1 = upper_bound(proj1.:C1) by Th46;
A3: E-bound C2 = upper_bound(proj1.:C2) by Th46;
  thus E-bound S = upper_bound(proj1.:S) by Th46
    .= upper_bound(proj1.:C1 \/ proj1.:C2) by A1,RELAT_1:120
    .= max(E-bound C1, E-bound C2) by A2,A3,SEQ_4:143;
end;
