Proving from De Morgan's laws

Question

I was proving:

$J.K'.L+J.K.L'+JKL=J.(L+K)$

I have done the following:

From LHS: $J.K'.L+J.K.L'+JKL$

$=J.(K'.L+K.L'+K.L)$

$=J.(K'.L+K.(L'+L))$

Since $L'+L=1:$

$=J.(K'L+K)$ [' is complement]

Now I'm unable to get the expression towards $X=J.(L+K)$

Could you please suggest me how I can prove this problem?

Answer 1

Nice progress. We will prove the identity $$A+B=A+A'B$$ for general variables $A,B$. You can use a truth table to see that this is indeed true, but we will prove it algebraicly.

Start with $$A+B$$ Since $A+A'=1$, this is equivalent to $$=1(A+B)$$ $$=(A+A')(A+B)$$ Expanding, yields $$=A+AB+AA'+A'B$$ $$=A(1+B)+1+A'B$$ $$=A+A'B$$

Hence, we have $$J(K'L+K)=J(L+K)$$

Answer 2

Distribution says: $P+QR=(P+Q)(P+R)$

So:

$K'L+K = (K'+K)(L+K) = 1(L+K)= L+K$