theorem
  g tolerates h & g in sproduct f & h in sproduct f
  implies g \/ h in sproduct f
proof
  assume that
A1: g tolerates h and
A2: g in sproduct f and
A3: h in sproduct f;
A4: {g,h} c= sproduct f by A2,A3,ZFMISC_1:32;
A5: now
    let h1,h2 be Function;
    assume that
A6: h1 in {g,h} and
A7: h2 in {g,h};
A8: h1 = g or h1 = h by A6,TARSKI:def 2;
    h2 = g or h2 = h by A7,TARSKI:def 2;
    hence h1 tolerates h2 by A1,A8;
  end;
  union {g,h} = g \/ h by ZFMISC_1:75;
  hence thesis by A4,A5,Th62;
end;
