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I need to simplify a Boolean algebraic equation and name the laws I use at each step, here is where I am at

AB + AC + B

I got the answer online but couldn't identity the law

AB + AC + B
(AB + B) + AC (Is this the commutative law?)
B + AC   (Name of this law?)
Ninja2k
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    In the first case, you have made use of the commutative law for $+$, (AC +B = B+AC), and also, in a sense you use the associative law as well: $AB+B +AC = (AB+ B) + C = AB +(B+C)$ – amWhy Apr 29 '17 at 15:44
  • $AB+ B = B$ is the absorption law, which holds in all lattices. – Fabio Somenzi Apr 29 '17 at 16:43
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    Yes @Fabio, I the use of the "absorption law" in "part II" of my answer below. Never hurts, though, to know why the absorption law works, and the laws/properties, identities on which it depends. I've always found a firm knowledge base and understanding some basic laws and properties, and then using them as building blocks, to reach the laws that depend on them to be empowering. – amWhy Apr 29 '17 at 16:58
  • @amWhy I agree. – Fabio Somenzi Apr 29 '17 at 17:07

2 Answers2

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In the first case, (second line following the proposition), you have made use of the commutative law for $+$, $$(AC + B = B +AC)\tag{associativity}$$ and also, the associative law for $+$ is being used: $$\color{blue}{(AB+B +AC)= ((AB+ B) + AC)} = (AB +(B+AC))\tag{commutativity}$$

$$ $$

To get to the second line $\large ^\left(\color{red}\dagger\right)$, we note that $B = 1\cdot B$, by the identity for "and". Then we can use the distributive property of "and" over "or" to get $$(AB+B) = AB+ 1\cdot B = (A+1)B\tag{distributivity}$$

Now, $$A+1 = 1\tag{annihilator axiom for +}$$ so we can reduce $$(A+1)B = 1\cdot B = B\tag{identity: $1\cdot B = B$}$$

In the end, we have $$AB + B + AC = B+AC$$

$$ $$ $(\color{red} \dagger)$ By invoking the axiom of absorption for "or" we can immediately conclude: $$AB+B = B$$ giving us $$((AB+B)+AC) = B+AC$$

You might want to bookmark the following link: Justifications(laws) of Boolean Logic. There you'll find references to each reason listed above.

amWhy
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Let's go step by step: your three term addition is actually $(AB+AC)+B$, so \begin{align} (AB+AC)+B &=AB+(AC+B) && \text{associativity} \\ &=AB+(B+AC) && \text{commutativity} \\ &=(AB+B)+AC && \text{associativity} \\ &=(AB+1B)+AC && \text{by $B=1B$} \\ &=\bigl((A+1)B\bigr)+AC && \text{distributivity} \\ &=1B+AC && \text{by $A+1=1$} \\ &=B+AC && \text{by $B=1B$} \end{align} The steps marked “associativity” are usually not mentioned. Also $$ AB+B=B $$ follows from $AB\le B$, so a quicker derivation is $$ AB+AC+B=AB+B+AC=B+AC $$ where for the first equality we apply the commutative law.

egreg
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