theorem
  a is dualizing implies b -r> c = (b [*] (c -l> a)) -r> a & b -l> c = (
  (c -r> a) [*] b) -l> a
proof
  deffunc F(set) = $1;
  defpred P1[Element of Q] means $1 [*] b [= c;
  defpred P2[Element of Q] means $1 [*] (b [*] (c -l> a)) [= a;
  defpred P3[Element of Q] means b [*] $1 [= c;
  defpred P4[Element of Q] means (c -r> a) [*] b [*] $1 [= a;
  assume
A1: (d -r> a) -l> a = d & (d -l> a) -r> a = d;
  then
A2: c = (c -l> a) -r> a;
A3: P1[d] iff P2[d]
  proof
    d [*] (b [*] (c -l> a)) = d [*] b [*] (c -l> a) by GROUP_1:def 3;
    hence thesis by A2,Th12;
  end;
  {F(d1): P1[d1]} = {F(d2): P2[d2]} from FRAENKEL:sch 3(A3);
  hence b -r> c = (b [*] (c -l> a)) -r> a;
A4: c = (c -r> a) -l> a by A1;
A5: P3[d] iff P4[d]
  proof
    (c -r> a) [*] b [*] d = (c -r> a) [*] (b [*] d) by GROUP_1:def 3;
    hence thesis by A4,Th11;
  end;
  {F(d1): P3[d1]} = {F(d2): P4[d2]} from FRAENKEL:sch 3(A5);
  hence thesis;
end;
