Intuition about a logic equivalence

Question

It is fairly easy to show that $(P\wedge Q)\to R$ is equivalent to $(P\to R)\lor(Q\to R)$.

However I am having trouble having an intuition about this equivalence. My intuition is that if both hypotheses $P$ and $Q$ are required to prove $R$, neither one by itself is sufficient to prove $R$.

Example 1: If I am very talented at basketball ($P$) and I am a man (Q) then I have a chance to play in NBA ($R$).

Example 2: If $p>2$ and $p$ is prime then $p$ is odd. However neither hypotheses by themselves imply that $p$ is an odd number.

Certainly my intuition is very wrong, could anyone give me perhaps more insight?

Answer 1

The confusion here isn't with implication, but with your implicit use of a universal quantifier.

Look at your Example 2, but write out the quantifier (I'll assume that quantifiers are taken to quantify over the natural numbers):

$$(\forall p)\big((p\gt 2 \,\wedge\, p\text{ is prime}) \to p\text{ is odd}\big). \tag{1}$$

This means that every natural number that is both greater than 2 and prime is, in fact, odd. (This is true, of course.)

By your propositional logic equivalence, it's correct to say that (1) is equivalent to:

$$(\forall p)\big((p\gt 2 \to p\text{ is odd}) \lor (p\text{ is prime }\to \;p\text{ is odd})\big),\tag{2}$$

although (2) is a strange, hard-to-interpret statement that nobody would write in that way.

This is not, however, equivalent to:

$$\big((\forall p)(p\gt 2 \to p\text{ is odd})\big)\lor\big((\forall p)(p\text{ is prime} \to p\text{ is odd})\big)\tag{3}$$

nor is it equivalent to

$$(\forall p)\big((p\gt 2 \,\lor\, p\text{ is prime} )\to p\text{ is odd})\big).\tag{4}$$

(3) means: Either (a) every natural number greater than 2 is odd or (b) every prime number is odd (or both).

(4) means: Every natural number that is either greater than 2 or prime (or both) is odd.

Both (3) and (4) are false — and, more to the point, they clearly aren't equivalent to (1) and to (2), nor does the propositional logic equivalence say that they should be.

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So, in propositional calculus it's true that $(P\wedge Q)\to R$ is equivalent to $(P\to R)\lor(Q\to R),$ and it's therefore true in first-order logic that

$$(\forall x)\big((P(x) \wedge Q(x))\to R(x)\big)$$

is equivalent to

$$(\forall x)\big((P(x)\to R(x))\lor(Q(x)\to R(x)) \big).$$

But this isn't equivalent to $$(\forall x)\big(P(x)\to R(x)\big) \lor (\forall x)\big(Q(x)\to R(x)\big),$$ for example. In general, $(\forall x)(A(x)\lor B(x))$ isn't equivalent to $(\forall x)A(x) \lor (\forall x)B(x).$

Answer 2

Mitchell Spector's answer explained very well what's wrong with your examples. Let me try to explain intuitively why $(P\land Q)\to R$ is equivalent to $(P\to R)\lor(Q\to R).$ (Since this is an intuitive explanation, I feel absolved from any need to be rigorous or correct.)

Let me identify a proposition with its truth value, i.e., $P=1$ if $P$ is true, $P=0$ if $P$ is false. Then $P\to R$ is true just in case $P\le R,$ i.e., "$R$ is at least as true as $P$ is." Since $P\land Q=\min(P,Q),$ the equivalence $$(P\land Q)\to R\iff(P\to R)\lor(Q\to R)$$ just says that $$\min(P,Q)\le R\iff P\le R\text{ or }Q\le R$$ which is a simple arithmetical fact, nothing unintuitive about it.