Let $a$, $b$ and $c$ be elements in a boolean algebra and $a+c = a+b$ and $ab = ac$. Then prove that $b = c$.

Question

To prove it, I did the following:

We suppose two different values for $a$; $a = 0$ and $a \not= 0$

1. if $a=0$ then eliminating two $a$ form $a+c=a+b$ leads us to $b=c$

2. if $a \not= 0$ then from $ab=ac$ we have:

$ab-ac=0$ ------> $a(b-c)=0$ --- $a\not=0$ ---> $b-c=0$ ------> $b=c$

Is my proof correct and acceptable?

Answer 1

You should be acquainted with distributivity and absorption.
With those properties, if $$a + c = a + b$$ and $$ab = ac,$$ then \begin{align} b &= b(a+b)\\ &= b(a+c)\\ &= ab + bc\\ &= ac + bc\\ &= (a + b)c\\ &= (a + c)c\\ &= c. \end{align} (I'll leave it to you to identify which property is used where.)