How to distribute $(\neg \lor ) \land [((\neg \lor ) \land ( \lor \neg ))∨((\neg \lor ) \land (\lor \neg ))]?$

Question

I see two possible ways to distribute this.

I believe it is case 1.

For brevity,

Let \begin{align}A &= (\neg P \lor Q) & B &= (P \lor \neg Q) & C &=(\neg R \lor Q) & D &=(R \lor \neg Q). \end{align}

Case 1:

$$(\neg P \lor R) \land [(A\land B)\lor (C\land D)]=$$

$$[(\neg P\land A \land B)\lor(\neg P \land C \land D)]\lor [(R \land A \land B)\lor (R \land C \land D)]$$

Case 2:

$$(\neg P \lor R)\land [(A \land B)\lor (C\land D)]=$$

$$[(\neg P \lor A) \land (\neg P \land B)\lor(\neg \land C)∧(\neg \land D]\lor [(R \land A)\land (R \land B)) \lor (R \land C) \land (R \land D)]$$

Question: Which case is correct (if either)? Why is this case correct/other wrong?

Answer 1

Case 1 is correct.

What you end up with in case 2 is ungrammatical ... you'll need to add some parentheses. I assume you meant:

$$[\color{red}((\neg P \color{red}\land A) \land (\neg P \land B)\color{red})\lor\color{red}((\neg \land C)∧(\neg \land D\color{red})\color{red})]\lor [\color{red}((R \land A)\land (R \land B)) \lor \color{red}((R \land C) \land (R \land D)\color{red})]$$

But note that that is equivalent to:

$$[((\neg P \land A) \land (\neg P \land B))\lor((\neg \land C)\land(\neg \land D))]\lor [((R \land A)\land (R \land B)) \lor ((R \land C) \land (R \land D))] \overset{Association \ (drop \ parentheses \ in \ generalized \ conjunctions)}\Leftrightarrow$$

$$[(\neg P \land A \land \neg P \land B)\lor(\neg \land C\land \neg \land D)]\lor [(R \land A\land R \land B) \lor (R \land C \land R \land D)] \overset{Commutation}\Leftrightarrow$$

$$[(\neg P \land \neg P \land A \land B)\lor(\neg \land \neg \land C \land D)]\lor [(R \land R \land A \land B) \lor (R \land R \land C \land D)] \overset{Idempotence}\Leftrightarrow$$

$$[(\neg P \land A \land B)\lor(\neg \land C \land D)]\lor [(R \land A \land B) \lor (R \land C \land D)]$$

... which is exactly what you got from case 1

In other words, had you used proper parentheses, both results would have been correct!

That said, I have a feeling you did the following to get to case 2:

$$(\neg P \lor R) \land [(A\land B)\lor (C\land D)]=$$

$$[(\neg P \land ((A\land B)\lor (C\land D))] \lor [(R \land ((A\land B)\lor (C\land D))]=$$

$$[(\neg P \land (A\land B)) \lor (\neg P \land (C\land D))] \lor [(R \land (A\land B))\lor (R \land (C\land D))]=$$

$$[(\neg P \land A) \land (\neg P \land B)) \lor (\neg P \land C) \land (\neg P \land D))] \lor [(R \land A) \land (R \land B))\lor (R \land C) \land (R \land D))]$$

That is, you acted as if that last step was a Distribution ... but it isn't! ... because you end up distributing a $\land$ over a $\land$. Distribution is $\land$ over $\lor$ or vice versa.

So, you should have just stopped 1 line earlier, and thus have ended up with

$$[(\neg P \land (A\land B)) \lor (\neg P \land (C\land D))] \lor [(R \land (A\land B))\lor (R \land (C\land D))]$$

but since $\land$ is Associative you can drop some parantheses and thus get:

$$[(\neg P \land A \land B)\lor(\neg \land C \land D)]\lor [(R \land A \land B) \lor (R \land C \land D)]$$

and since $\lor$ is Associative you can drop those square brackets as well:

$$(\neg P \land A \land B)\lor(\neg \land C \land D)\lor (R \land A \land B) \lor (R \land C \land D)$$

Answer 2

Case 1 is right. Case 2 is not even a formula since some parentheses are missing: in $(\lnot P \lor A) \land (\lnot P \land B) \lor (\lnot P \lor C) \land (\lnot P \lor D)$ it is not clear what is the principal connective.

Let us see why case 1 is correct. Essentially, you use repeatedly the following distributive law (together with associativity and commutativity): \begin{align} A \land (B \lor C) &\equiv (A \land B) \lor (A \land C) \end{align}

By applying step-by-step this equivalence rule, you get: \begin{align} &(\neg P \lor R) \color{red}{\land} \big((A\land B) \color{red}{\lor} (C\land D)\big) \\ \text{(distributivity) } \ \equiv \ & ((\neg P \lor R) \land (A \land B)) \lor ((\neg P \lor R) \land (C \land D)) \\ \text{(commutativity) } \ \equiv \ & ((A \land B) \color{red}{\land} (\neg P \color{red}{\lor} R)) \lor ((C \land D) \color{red}{\land} (\neg P \color{red}{\lor} R)) \\ \text{(distributivity) } \ \equiv \ & \big(((A \land B) \land \neg P) \lor ((A \land B) \land R)\big) \lor \big(((C \land D) \land \lnot P) \lor ((C \land D) \land R)\big) \\ \text{(associativity) } \ \equiv \ & \big((A \land B \land \neg P) \lor (A \land B \land R)\big) \lor \big((C \land D \land \lnot P) \lor (C \land D \land R)\big) \end{align}