Can $(A \lor B) \land (\lnot A \land \lnot C)$ be more simplified?

Question

Can $(A \lor B) \land (\lnot A \land \lnot C)$ be more simplified/expanded? With a kind of distributive property?

Answer 1

$$\begin{align} (A\lor B)\land (\lnot A \land \lnot C) & \equiv [(A\lor B) \land \lnot A] \land \lnot C \\ \\ & \equiv [\underbrace{(A\land \lnot A)}_{\text{False}}\lor (\lnot A \land B)] \land \lnot C \\ \\ & \equiv \lnot A \land B \land \lnot C \end{align}$$

The first equivalence is simply due to the associativity of $\land$.

The second is by use of the distributive property, distributing a conjunct over a disjunction.

Third, we have $F\lor P\equiv P$.

Answer 2

$(A \lor B) \land (\lnot A \land \lnot C) \longleftrightarrow (A \land \lnot A \land \lnot C) \lor (B \land \lnot A \land \lnot C) \longleftrightarrow False \lor (B \land \lnot A \land \lnot C) \longleftrightarrow B \land \lnot A \land \lnot C$

Answer 3

We can "dsitribute" the OR in $(A \vee B) \wedge (\neg A \wedge \neg C)$ to get
$[A \wedge (\neg A \wedge \neg C)] \vee [B \wedge (\neg A \wedge \neg C)]$

One of the terms should seem supicious.

Answer 4

$ \newcommand{\calc}{\begin{align} \quad &} \newcommand{\calcop}[2]{\\ #1 \quad & \quad \unicode{x201c}\text{#2}\unicode{x201d} \\ \quad & } \newcommand{\endcalc}{\end{align}} \newcommand{\ref}[1]{\text{(#1)}} \newcommand{\iff}{\equiv} \newcommand{\true}{\text{true}} \newcommand{\false}{\text{false}} $One law of logic that is not very well known, but very useful in cases like this one, is $$ \tag 0 P \land E(P) \;\iff\; P \land E(\true) $$ ("assume one side of $\;\land\;$ is $\;\true\;$ on the other side") together with its dual $$ \tag 1 P \lor E(P) \;\iff\; P \lor E(\false) $$ where $\;E(P)\;$ is any boolean expression which contains $\;P\;$ as a free variable.

This allows us to do the simplification even more quickly than in the earlier answers:

$$\calc (A \lor B) \land \lnot A \land \lnot C \calcop\iff{by $\ref 0$: assume $\;\lnot A\;$ is $\;\true\;$ on the left hand side of the first $\;\land\;$} (\false \lor B) \land \lnot A \land \lnot C \calcop\iff{simplify: $\;\false\;$ is the identity of $\;\lor\;$} B \land \lnot A \land \lnot C \endcalc$$

Of course, to prove $\ref 0$ and $\ref 1$ you still would need to use distribution and contradiction as in the older answers, together with induction over the structure of boolean expressions... But when you're using logic in calculations, then these generalized laws often come in very useful.