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I’m given a question and its to prove exclusive OR can be the equivalent of using a NAND, OR and a AND gate show in the Boolean equation.

$$A \oplus B = (A+B) \cdot (\overline{AB})$$

But when I look at the solutions, loads of steps have been missed and it doesn’t make sense to me.

Solution

I only understand the first line and it looks like on the second line, terms were added but theres no indication in what rule was used to this. I've tried it on my own and I tried 'adding terms' but I just got some long equation that couldn't be simplified

I was wondering if anyone could give guidance on how this proof was done.

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    \$\begingroup\$ What "doesn't make sense"? What line? What have you worked out on your own? Where are you stuck? Do you understand the first line? Do you see how to get from the first to the second? ??? \$\endgroup\$ – scorpdaddy Apr 24 at 14:46
  • \$\begingroup\$ Taking the last line and expanding it to work back towards the start might make the equivalence clearer. \$\endgroup\$ – Phil G Apr 24 at 14:50
  • \$\begingroup\$ Boolean Algebra expands to contract as required. In step 2, you just Add 0. \$\endgroup\$ – StainlessSteelRat Apr 24 at 16:24
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Is there a particular step that you find unclear? The steps can be explained using the laws of Boolean Algebra.

To go from line 1 to line 2, it simply adds terms (using Complement Law) for

$$A*!A$$

and

$$B*!B$$

which are both clearly zero.

Since

$$X + 0 = X$$

then line 2 and line 1 are equal.

Line 2 to line 3 is factoring out common terms (using Distributive Law); i.e.

$$A * X + A * Y = A * (X + Y)$$

Line 3 to 4 is applying DeMorgan's theorem:

$$!(AB) = (!A + !B)$$

Line 4 to 5 is once again factoring out the common term (using Distributive Law) $$!(AB)$$

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