Your first job is to do a truth table for this sentence: Pv(Q&R).

Use a new piece of paper, because you’ll need the reference columns for more sentences.

Next, figure out which of these sentences has the same truth function. You can use the same reference columns you’ve already created.

Congratulations: you’ve discovered another key equivalence in BOOL.

**Distribution (Dist):**

Pv(Q&R) ⇔ (PvQ)&(PvR)

P&(QvR) ⇔ (P&Q)v(P&R)

This equivalence shows how to distribute v over &. The same principle applies for distributing & over v.

These principles are similar to DeMorgan’s, since they show how the different connectives interact. But there are some important differences to keep in mind.

Distribution changes the number of atomics in the sentence, but DeMorgan’s doesn’t.

Additionally, DeMorgan’s is best thought of as “flipping” the v to & or vice versa.

~(PvQ) becomes ~P&~Q.

But we recommend you don’t think of distribution in the same way. When you use distribution on Pv(Q&R), think of “Pv” as a chunk that you put on Q and put on R, like this:

That creates (PvQ) and (PvR).

Then the & that was originally between the Q and R becomes the new main connective: (PvQ)&(PvR).

In distribution, then, neither connective flips; they just change positions.

It’s easier to remember how distribution works when you realize it is just like laws in algebra. Since BOOL has structurally similar laws, the logic is often called a Boolean algebra.

For example, when we have 2×(3+4), we treat “2×” as a chunk, and put it on the 3 and the 4: (2×3)+(2×4). Then the + becomes the new main operator.

Distribution in BOOL also works if the main connective is on the left side, like in the next problem.

If a sentence is more complex, you might need to perform distribution multiple times. Take this example:

(A&B)v(C&D)

It is structurally similar to Pv(Q&R). Q is replaced by C; R is replaced by D; and P is replaced by the complex sentence (A&B).

Distribution works the same way. We treated “Pv” as a chunk, so now treat “(A&B)v” as a chunk.

We distributed the disjunction onto the conjunction. But since the sentence was more complex and had two conjunctions, we can keep going.

In the result ((A&B)vC)&((A&B)vD), we can now treat “vC” and “vD” as chunks, and distribute them in the right places.

Here’s the final result:

(AvC)&(BvC)&(AvD)&(BvD)

Now we are done distributing the disjunction, because we don’t have any more disjunctions wide scope around conjunctions.

You can use a truth table to convince yourself that that sentence really is equivalent to the original.

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