theorem Th35:
  b in B & c in C & b \/ c in DISJOINT_PAIRS A implies b \/ c in B ^C
proof
  assume b in B & c in C;
  then
A1: b \/ c in { s \/ t: s in B & t in C };
  assume b \/ c in DISJOINT_PAIRS A;
  hence thesis by A1,XBOOLE_0:def 4;
end;
