theorem Th142:
  for a,b being set holds (a .--> b)*(n|->a) = n |-> b
proof
  let a,b be set;
A1: now
    let x be object;
    hereby
      assume x in dom (n |-> b);
      then
A2:   x in Seg n;
      hence x in dom(n|->a);
      dom(a .--> b) = {a} & (n|->a).x = a by A2,FUNCOP_1:7;
      hence (n|->a).x in dom(a .--> b) by TARSKI:def 1;
    end;
    assume that
A3: x in dom(n|->a) and
    (n|->a).x in dom(a .--> b);
    x in Seg n by A3;
    hence x in dom (n |-> b);
  end;
  now
    let x be object;
A4: a in {a} by TARSKI:def 1;
    assume x in dom (n |-> b);
    then
A5: x in Seg n;
    hence (n |-> b).x = b by FUNCOP_1:7
      .= (a .--> b).a by A4,FUNCOP_1:7
      .= (a .--> b).((n|->a).x) by A5,FUNCOP_1:7;
  end;
  hence thesis by A1,FUNCT_1:10;
end;
