theorem Th1:
  dom f = dom g & (for c st c in dom f holds f/.c = g/.c) implies f = g
proof
  assume that
A1: dom f = dom g and
A2: for c st c in dom f holds f/.c = g/.c;
  now
    let x be object;
    assume
A3: x in dom f;
    then reconsider y=x as Element of C;
    f/.y=g/.y by A2,A3;
    then (f qua Function).y = g/.y by A3,PARTFUN1:def 6;
    hence (f qua Function).x = (g qua Function).x by A1,A3,PARTFUN1:def 6;
  end;
  hence thesis by A1;
end;
