theorem Th79:
  for E1 be Enumeration of F1,E2 be Enumeration of F2 st F1 misses F2
    holds E1^E2 is Enumeration of F1\/F2
proof
  let E1 be Enumeration of F1,E2 be Enumeration of F2
    such that
A1: F1 misses F2;
  rng E1=F1 & rng E2=F2 by RLAFFIN3:def 1;
  then E1^E2 is one-to-one&rng (E1^E2) = F1\/F2 by A1,FINSEQ_3:91,FINSEQ_1:31;
  hence thesis by RLAFFIN3:def 1;
end;
