(AB)'(AB) = I (B'A')(AB) = [(B'A')A]B = [B'(A'A)]B = [B'I]B = B'B = I = (AB)'(AB) So (B'A')(AB) = (AB)'(AB) Right multiply both sides by (AB)' [(B'A')(AB)](AB)' = [(AB)'(AB)](AB)' (B'A')[(AB)](AB)'] = (AB)'[(AB)(AB)'] (B'A')I = (AB)'I B'A' = (AB)' That's what we were to prove. Edwin