
theorem Th5:
  for S1,S2 being non empty ManySortedSign st S1 tolerates S2 holds
  InputVertices (S1+*S2) = ((InputVertices S1)\(InnerVertices S2)) \/ ((
  InputVertices S2)\(InnerVertices S1))
proof
  let S1,S2 be non empty ManySortedSign;
A1: ((the carrier of S1) \ ((rng the ResultSort of S1)\/(rng the ResultSort
  of S2))) =(InputVertices S1) \ (InnerVertices S2) by XBOOLE_1:41;
  assume S1 tolerates S2;
  then
A2: the ResultSort of S1 tolerates the ResultSort of S2 by CIRCCOMB:def 1;
  InputVertices (S1+*S2) = (the carrier of S1+*S2) \ rng ((the ResultSort
  of S1) +* (the ResultSort of S2)) by CIRCCOMB:def 2
    .= (the carrier of S1)\/(the carrier of S2) \ rng ((the ResultSort of S1
  ) +* (the ResultSort of S2)) by CIRCCOMB:def 2
    .= (the carrier of S1)\/(the carrier of S2) \ ((rng the ResultSort of S1
  )\/rng the ResultSort of S2) by A2,FRECHET:35
    .= ((the carrier of S1) \ ((rng the ResultSort of S1)\/(rng the
ResultSort of S2)))\/ ((the carrier of S2) \ ((rng the ResultSort of S1)\/(rng
  the ResultSort of S2))) by XBOOLE_1:42;
  hence thesis by A1,XBOOLE_1:41;
end;
