theorem Th92:
  f.:A=f.:C implies A=C
proof
  assume
A1: f.:A=f.:C;
  now
    let b be object;
    assume
A2: b in C;
    then reconsider b9=b as Element of AFS;
    set y=f.b9;
    y in f.:C by A2,Th90;
    then ex a st a in A & f.a=y by A1,Th91;
    hence b in A by FUNCT_2:58;
  end;
  then
A3: C c= A by TARSKI:def 3;
  now
    let a be object;
    assume
A4: a in A;
    then reconsider a9=a as Element of AFS;
    set x=f.a9;
    x in f.:A by A4,Th90;
    then ex b st b in C & f.b=x by A1,Th91;
    hence a in C by FUNCT_2:58;
  end;
  then A c= C by TARSKI:def 3;
  hence thesis by A3,XBOOLE_0:def 10;
end;
