
theorem Th1:
  for X being non empty set holds
  {[A,B] where A,B is Element of X : (B = {} implies A = {})} =
  [:X,X:] \ [: X \ { {} } , { {} } :]
  proof
    let X be non empty set;
    hereby
      let o be object;
      assume o in {[A,B] where A,B is Element of X : (B = {} implies A = {})};
      then consider A,B be Element of X such that
A1:   o = [A,B] and
A2:   B = {} implies A = {};
A3:   o in [:X,X:] by A1,ZFMISC_1:def 2;
      now
        assume o in [: X \ { {} } , { {} } :];
        then ex a,b be object st a in X \ {{}} & b in { {} } & o = [a,b]
          by ZFMISC_1:def 2;
        hence contradiction by A1,A2,XTUPLE_0:1;
      end;
      hence o in [:X,X:] \ [: X \ { {} } , { {} } :] by A3,XBOOLE_0:def 5;
    end;
    let o be object;
    assume
A4: o in [:X,X:] \ [:X \ {{}} , {{}} :];
    then o in [:X,X:] & not o in [: X \ { {} } , { {} } :] by XBOOLE_0:def 5;
    then consider u,v be object such that
A5: u in X and
A6: v in X and
A7: o = [u,v] by ZFMISC_1:def 2;
    reconsider u,v as Element of X by A5,A6;
    not [u,v] in [: X \ {{}} , {{}} :] by A7,A4,XBOOLE_0:def 5;
    then not (u in X \ {{}} & v in {{}}) by ZFMISC_1:def 2;
    then (not (u in X & not u in {{}})) or (not v in {{}}) by XBOOLE_0:def 5;
    then (not u in X) or (u = {}) or (not v ={}) by TARSKI:def 1;
    hence o in {[A,B] where A,B is Element of X : (B = {} implies A = {})}
      by A5,A6,A7;
  end;
