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In the boolean logic

$ab + ac + \bar a \bar b c$

is equivalent to the simpler

$ab + \bar b c$

This can be confirmed by looking at the truth table (below).

The $ab$ and $\bar b c$ minterms cover the $ac$ minterm.

My question is, in general/practice, how can one make this simplification algebraically, that is without having to resort to looking at truth table, Karnaugh maps, which set of minterms possibly cover another minterm, etc.

I tried to manipulate $ab + ac + \bar b c$ algebraically, but only got as far as $a(b + c) + \bar b c\ $ or $\ ab + (a + \bar b)c$ and couldn't really simplify further from there.

Notation:

  • $ab$ means $a \wedge b$
  • $a + b$ means $a \vee b$
  • $\bar b$ means $\neg b$

Truth table:

a b c ab ac ¬b c ab + ac + ¬b c ab + ¬b c
0 0 0
0 0 1 1 1 1
0 1 0
0 1 1
1 0 0
1 0 1 1 1 1 1
1 1 0 1 1 1
1 1 1 1 1 1 1
Andrei
  • 37,370
joseville
  • 1,477

2 Answers2

2

Note: I use $x'$ rather than $\bar x$ for ease of writing.

Since the initial expression is ($ab+\cdots$) and the resulting one is also ($ab+\cdots$) let's focus first on the group $X=ac+a'b'c$

Seems factoring $c$ out is the way to go so let's do it $X=(a+a'b')c$

Now we would like to make $b'$ appear, we can replace $1=b+b'$ to get

$X=(a(b+b')+a'b')c = (ab+ab'+a'b')c = (ab+(a+a')b')c = (ab+b')c = abc+b'c$

Now the expression $abc$ is simply swallowed by $ab$ when calculating the resulting $ab+X$ because $abc+ab=ab(c+1)=ab(1)=ab$.

So maybe try yo remember the trick $1=x+x'$ it is quite handy in boolean logic.

zwim
  • 28,563
2

$$ab + ac + \bar{a}\bar{b}c$$ $$= ab + (a + \bar{a}\bar{b})c$$ $$= ab + (a(b + \bar{b}) + \bar{a}\bar{b})c$$ $$= ab + (ab + a\bar{b} + \bar{a}\bar{b})c$$ $$= ab + (ab + (a + \bar{a})\bar{b})c$$ $$= ab + (ab + \bar{b})c$$ $$= ab + abc + \bar{b}c$$ $$= ab(1 + c) + \bar{b}c$$ $$= ab + \bar{b}c$$

Dan
  • 14,978