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I have found these boolean expressions based on a truth table:

  1. $(abc'd') + (ab'c'd) + (a'bcd') + (a'b'cd)$

(EDITED for 2))

  1. $(abcd') + (abc'd) + (ab'cd) + (ab'c'd') + (a'bcd) + (a'bc'd') + (a'b'c'd) + (a'b'cd')$

However, I need to simplify them. (' for complement, + for or, xy is for x and y)

2 Answers2

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$1. ~~(abc'd')+(ab'c'd) \iff ac'(bd'+b'd),$ and $(a'bcd')+ (a'b'cd) \iff a'c(bd'+b'd)$, so this yields $(ac'+a'c)(bd'+b'd) \iff (a \oplus c)(b \oplus d)$.

$2.$ Factor as in problem $1$ using $ab'$ and $a'b$, and then $ab$ and $a'b'$, and you'll end up with $(a \oplus b) \oplus (c \oplus d)$.

Robert Shore
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  • @anonymouseeeee Still not sure it's right. Three terms start with $ab$ and only one starts with $a'b'$. That could be what's intended but I suspect more symmetry was meant. – Robert Shore Sep 22 '20 at 03:46
  • Yes, sorry about that. There was a mistake in my truth table. I have fixed it once more. –  Sep 22 '20 at 12:19
  • @anonymouseeeee The symmetry leads to a nice answer. – Robert Shore Sep 23 '20 at 01:53
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Using $\oplus$ to denote the XOR operation.

For the first one:

\begin{align} &(ab{c'}d') + ({a}b'{c'}d) + ({a'}b{c}d') + ({a'}b'{c}d)&\\\ &{ac'}{(bd'+b'd)}+{a'c}{(bd'+b'd)}&\text{Factor}\\ &ac'{(b\oplus d)}+a'c{(b\oplus d)}&\text{Use }\oplus\\ & {(b\oplus d)}(ac'+a'c)&\text{Factor}\\ &(b\oplus d)(a\oplus c)&\text{Use }\oplus \end{align}

For the second one:

\begin{align}&{ (abcd') + (abc'd) + (ab'cd) + (ab'c'd') + (a'bcd) + (a'bc'd') + (a'b'c'd) + (a'b'cd')}&\\ &{ab(cd'+c'd)+cd(ab'+a'b)+a'b'(cd'+c'd)+c'd'(ab'+a'b)}&\text{Factor}\\ &{ab(c\oplus d)+cd(a\oplus b)+a'b'(c\oplus d)+c'd'(a\oplus b)}&\text{Use }\oplus\\ &{(c\oplus d)(ab+a'b')+(a\oplus b)(cd+c'd')}&\text{Factor}\\ &{(c\oplus d)(a\oplus b)'+(a\oplus b)(c\oplus d)'}&\text{Use }\oplus\\ &{(c\oplus d)\oplus(a\oplus b)}&\text{Use }\oplus \end{align}

user400188
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