boolean-algebra
So I'm doing some homework and trying to simplify $(x'+y)'(x+y)'$. So far these are the steps I've completed, but I'm not 100% sure that they're appropriate.
$(x'+y)'(x+y)' = (x'+y)'(x’y’)$
$(x'+y)'(x’y’) = (x’’+ y’)(x’y’)$
$(x’’+y’)(x’y’) = (x+y’)(x’y’)$
I'm unsure of what to do after this step (or if I've made any actual mistakes)
Let me try this-- I have added two additional step in bold to make the steps clear.
$(a' + c' + b'·d')·(b' + c + d')·(a + b')$
$=\mathbf{(a' + c' + b'·d').(a + b')·(b' + c + d')}$ (by coomutativity of $.$)
$= (a'·b' + a·c' + b'·c' + \underline{a·b'·d' + b'·d'})·(b' + c + d')$ (multiplying and rearranging the first two factors; commutativity and associativity of $.$ and $+$, just like addition and multiplication of real number, being used )
$= (a'·b' + a·c' + b'·c' + \underline{b'·d'})·(b' + c + d')$ (using $\mathbf{x+x.y}=x:(1+y)=x.1\mathbf{=x}$ in the underlined)
= a'·b' + a·b'·c' + b'·c' + b'·d' + a'·b'·c + b'·c·d' + a'·b'·d' + a·c'·d' + b'·c'·d'+ b'·d' (multiplying and rearranging just like one step above; also $x.x=x$ has been used in all the 2-product terms)
$=\mathbf{a'.b'+\underline{b'.c'.a+b'.c'}+\underline{b'.d'+b'.d'.(a'+c+c')}+a.c'.d}$ (note that $b'.d'+b'.d'=b'.d'$ as $x+x=x\quad\forall x$; also the rearrangement is for the next step)
$= a'·b' + b'·c' + b'·d' + a·c'·d$, using the rule $x+x.y=x$ with $x=b'.c', y=a$ in the first underline and $x=b'.d', y=(a'+c+c')$ in the second underline.
Your original expression is a product of sums:
(A' + B + C') (A + D') (C + D')
If you apply the and for the first two sums, you get:
(A'A + A'D' + AB + BD') (C + D')
A'A cancels out to false. In conjunction with the third sum, we get:
A'CD' + A'D'D' + ABC + ABD' + BCD' + BD'D'
Applying D'D' = D' gives us:
A'CD' + A'D' + ABC + ABD' + BCD' + BD'
A'CD' is covered by A'D' and can thus be omitted.
The minimized sum of products (the original six terms are covered by just four terms):
A'D' + C'D' + BD' + ABC
The terms of the expression shown in a Karnaugh-Veitch map:
The diagram helps to visually grasp which term is covered by which larger term.
Best Answer
$(x'+y)'(x+y)'=(x'+y)'(x'y')=(x''y')(x'y')=(xy')(x'y')$
Can you do the last step?