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I have a truth table for an odd parity checker with three inputs, the parity bit input, and the error bit as output. $p=1$ if an even number of the three inputs are $1$, and $e=1$ if an even number of bits between the three inputs and $p$ are $1$.

x y z p | e
--------+--
0 0 0 0 | 1
0 0 0 1 | 0
0 0 1 0 | 0
0 0 1 1 | 1
0 1 0 0 | 0
0 1 0 1 | 1
1 0 0 0 | 0
1 0 0 1 | 1
0 1 1 0 | 1
0 1 1 1 | 0
1 0 1 0 | 1
1 0 1 1 | 0
1 1 0 0 | 1
1 1 0 1 | 0
1 1 1 0 | 0
1 1 1 1 | 1

I've found the algebraic form of this truth table to be below:

$$A'B'C'D' + A'B'CD + AB'CD' + AB'C'D + A'BCD' + AB'C'D + AB'CD' + ABC'D' + ABCD$$

Trying to simplify it, it seems I can get rid of four terms:

\begin{align} &A'B'C'D' + A'B'CD + AB'CD' + AB'C'D + A'BCD' + AB'C'D + AB'CD' + ABC'D' + ABCD \\ &= (A'B')(CD + C'D') + (AB)(CD + C'D') + A'BC'D + A'BCD' + AB'C'D + AB'CD' \\ &= (A'B')(1) + (AB)(1) + A'BC'D + A'BCD' + AB'C'D + AB'CD' \\ &= A'B' + AB + A'BC'D + A'BCD' + AB'C'D + AB'CD' \\ &= 1 + A'BC'D + A'BCD' + AB'C'D + AB'CD' \\ &= A'BC'D + A'BCD' + AB'C'D + AB'CD'\end{align}

Is this as far as I can simplify it and am I correct in my deletion of terms?

$$A'B' + AB = 1$$

This is what I've used to delete my terms more or less.

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  • $\begingroup$ If $A'B' +AB$ means $(\neg A\wedge \neg B)\vee (A \wedge B)$, then no that is not always true. $\endgroup$ – kccu Oct 23 '16 at 23:34
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The best way here would be forming the K-MAP and then finding the minimal expression for the above truth table.

enter image description here

$F(x,y,z,p) = \sum(0,3,5,6,9,10,12,15)$ and it is not further reducible.

$F = \overline{x}.\overline{y}.\overline{z}.\overline{p} + \overline{x}.\overline{y}zp + \overline{x}y\overline{x}.\overline{y} + \overline{x}yz\overline{p} + xy\overline{z}.\overline{p} + xyzp + x\overline{y}.\overline{z}p + x\overline{y}z\overline{p}$

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