reserve x,y,z for set;
reserve f,f1,f2,f3 for FinSequence,
  p,p1,p2,p3 for set,
  i,k for Nat;

theorem Th35:
  x in rng f & y in rng(f-|x) implies f-|x-|y = f-|y
proof
  assume that
A1: x in rng f and
A2: y in rng(f-|x);
  thus f-|y = ((f -| x) ^ <* x *> ^ (f |-- x))-|y by A1,FINSEQ_4:51
    .= ((f -| x) ^ (<* x *> ^ (f |-- x)))-|y by FINSEQ_1:32
    .= f-|x-|y by A2,Th12;
end;
