Proof that $a + 1 = 1$ (Boolean algebra)

Question

The proof works like this:

$$a + 1 = (a + 1) * 1 = (a + 1) * (a + \lnot a) = a + (1 * \lnot a) = a + \lnot a = 1$$

It only uses the axioms of Boolean algebra. I understand every step besides the third one. He uses the distributive law here but I simply don't see how.

$$(\color{blue}{ a +} 1) (\color{blue}{ a +} \lnot a) = \color{blue}{ a +}(1\lnot a)$$ — amWhy, Oct 29 '18 at 18:19

score 2 · Accepted Answer · answered Oct 29 '18 at 18:18

2

There are two distributive laws in Boolean algebra (which are De Morgan duals of each other). The one in use here is $$(x+y)(x+z)=x+yz $$ with $x=a$, $y=1$, $z=\neg a$.

answered Oct 29 '18 at 18:18

hmakholm left over Monica

286,031

score 0 · Answer 2 · answered Oct 29 '18 at 19:14

0

by using this property $a+a'=1 $

$a+1=a+(a+a')=a+a'=1$

answered Oct 29 '18 at 19:14

Why is $a + (a+ a') = a + a'$? – Julian Oct 29 '18 at 19:21
in boolean $(a+a=a)$ not $2a$ – Oct 30 '18 at 02:54

Proof that $a + 1 = 1$ (Boolean algebra)

2 Answers2