Given:  P(AᑎB) = P(A)P(B)
Prove:  P(AᑎB') ≟ P(A)P(B') 

Let P(AᑎB')=w
Let P(AᑎB)=x
Let P(A'ᑎB)=y
Let P(A'ᑎB')=z

Then x+y+w+z = 1

Draw a Venn diagram with the probability of each region written
in each region:




     P(AᑎB) = P(A)P(B)      given
          x = (w+x)(x+y)    
 x(w+x+y+z) = (w+x)(x+y)    since w+x+y+z = 1
wx+x²+xy+xz = wx+wy+x²+xy   multiplying out
         xz = wy

We want to prove:

    P(AᑎB') ≟ P(A)P(B')
          w ≟ (w+x)(w+z)
 w(w+x+y+z) ≟ (w+x)(w+z)    since w+x+y+z = 1  
w²+wx+wy+wz ≟ w²+wz+wx+xz

So to prove that, we take

         wy = xz            same as xz = wy , proved above

                            and add w²+wx+wz to both sides:

wy+w²+wx+wz = xz+w²+wx+wz   
w²+wx+wy+wz = w²+wz+wx+xz   rearrange to reverse the above steps
 w(w+x+y+z) = (w+x)(w+z)
          w = (w+x)(w+z)    since w+x+y+z = 1
    P(AᑎB') = P(A)P(B')     which is what we had to prove.

Edwin