Without Truth Tables, how can you intuit that $P \implies Q \; \equiv \; \lnot P \vee Q $, and $ \not \equiv \; P \vee \lnot Q $?

Question

As I forgot to stipulate the pretermission of the logical equivalence $P → Q ≡ ¬P ∨ Q$ here, some of the answers presume this and resemble petitio principii, and so are really proofs rather than intuitive explanations. So for this question: exclude, and do NOT use or rely on, any foreknowledge of ¬P ∨ Q as the answer, formal proofs, Substitution Instances or Truth Tables. Before attempting any proofs, how would you conjecture or divine that P → Q:

1. $\equiv \quad \lnot P \vee Q $ and
2. $\not \equiv \quad P \vee \lnot Q $?

I do not understand some of this post that appears to explain the conjecture, but that I rewrote without Substitution Instances:

[3.] $\color{green}{\lnot P \vee Q}$ leaves open the possibility that $ P \wedge Q$.
[4.] In fact, $\color{green}{\lnot P \vee Q}$ says nothing about what happens when $P$.

[5.] $\color{#C41E3A}{ P ∨ ¬Q }$ states that $\lnot P$ AND $Q$ is impossible.

5. How does 3 help to decide between 1 and 2? If I understand correctly, 3 is true because 4 implies that both $P \wedge Q$ and $P \wedge \lnot Q$ are possible.

6. How does 5 help? I understand 5 via $\color{#C41E3A}{ P ∨ ¬Q } \quad \equiv \quad \lnot[\lnot P \wedge Q]$.

Answer 1

Consider the act of assertion :

An assertion is a speech act in which something is claimed to hold.

Thus, according to the "meaning" of $\lor$ (logical or, i.e. disjunction) that in order to assert the disjunction, at least one of the disjuncts must hold, when we assert :

$\lnot P \lor Q$,

we are claiming that :

"it is not the case that $P$ holds and $Q$ does not".

This is the "classical" expression for the so-called :

necessary condition :

the assertion that $Q$ is necessary for $P$ is colloquially equivalent to "$P$ cannot be true unless $Q$ is true".

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The presence of the negation is what "breaks the symmetry" between $\lnot P \lor Q$ and $P \lor \lnot Q$ and this account for the asymmetry between "necessary condition" and "sufficient condition".

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Note : in the "elucidation" of the ground for asserting $\lnot P \lor Q$ above ("it is not the case that ...") I'm not using any truth-functional properties of the logical connectives. I'm relying only on the "natural language" use of negation and conjunction, that is quite uncontroversial.

Answer 2

This will be bizarre (though technically grounded!) but is, actually, how I remember/think of it.

You can provide a computational semantics for classical propositional logic, and in that semantics negation behaves like what programmers call a continuation. So $\lnot A$ is the type of $A$ accepting continuations. Less esoterically, implication gets mapped to a function type. $A \Rightarrow B$ is the type of functions $A$ to $B$. Similarly, $A\land B$ is the type of pairs of $A$ and $B$, i.e. $A\times B$, and $A\lor B$ is the disjoint union of $A$ and $B$.

Clearly, a function $A \Rightarrow B$ accepts an $A$ and produces a $B$. In intuitionistic propositional logic, $\Rightarrow$ is usually primitive. But, as is well known, it's definable in classical propositional logic in terms of negation and conjunction or negation and disjunction. The more natural expansion computationally is $$A \Rightarrow B \equiv \lnot(A\land \lnot B)$$ This describes a continuation accepting a pair of a value of type $A$ and a continuation accepting a value of type $B$. This actually perfectly corresponds with typical assembly-level calling conventions for functions: the passed in $\lnot B$ is like a return address. The alternate definition $$A \Rightarrow B \equiv \lnot A \lor B$$ is less obvious to interpret computationally, but it still clearly communicates the fact that a function $A \to B$ will either consume an $A$ or produce a $B$. The fact that it may need to consume that $A$ to produce the $B$ is handled from the quite amusing computational interpretation of the equivalence $$\lnot(A\land\lnot B)\Leftrightarrow(\lnot A\lor B)$$ If you're a programmer, figuring it out is an enjoyable exercise. If you're not, then it wouldn't have much meaning anyway.

Answer 3

One way to approach $P\to Q$ is to ask ourselves what we would expect to be the case if someone asserted, If it gets cloudy then it will rain.

On the side, we may let $P$ be "it is cloudy" and $Q$ be "it rains", but first consider what we expect when we hear that statement. There are four possibilities. We only care about the first two of them:

1. It is cloudy and it rains.
2. It is cloudy and it does not rain.
3. It is not cloudy and it rains.
4. It is not cloudy and it does not rain.

If we accept what the person said as true, then we expect the first possibility to happen. However, if the second happens, then we tell the person he was wrong. But if the last two happen, that is, if it does not get cloudy at all, we neither praise nor blame the person for good or bad information.

Now replace the sentences with the symbols using an "and" connective ($\land$):

1. $P \land Q$: If this happens then the person spoke the truth.
2. $P \land \neg Q$: If this happens then the person spoke a falsehood.
3. $\neg P \land Q$: We don't care about this case since it did not get cloudy.
4. $\neg P \land \neg Q$: We don't care about this case either since it did not get cloudy.

If we want to build a bivalent truth-functional logic we will need to do two things.

First, we have to assign a truth value to any well-formed proposition. Since this is a bivalent system we can only assign true or false values. There is no third option and any well-formed proposition has to be assigned one or the other value.

Second, any connective such as "and", "or", "not" and "implies" has to be assigned a truth value also for all possible combinations of these well-formed propositions. That is where the truth tables come in. They show all these assigned values of true or false.

This second condition raises a problem. We really don't care what truth-value we assign to the 3rd and 4th possibilities above, but we have to assign both of them some truth value. Since we don't care, it is convenient to assign them the value true. Although that might appear to be a reasonable default under the circumstances of bivalence, that's what makes this appear non-intuitive in other contexts.

Now let's go to the questions:

• How can we make intuitive sense of $\neg P \lor Q$ representing $P \to Q$?

The only time when the conditional is false is when it is cloudy and it does not rain, or $P\land \neg Q$. Suppose it is not cloudy. Then it is one of the two situations where we don't care and have assigned that true. Suppose it is cloudy. We do care about these situations. If it is cloudy then it better rain.

So, $\neg P \lor Q$, it is not cloudy or it rains, represents what we expect (with the don't-care situations taken as true) for $P\to Q$, if it is cloudy then it rains.

• Why does $P \lor \neg Q$ not represent the conditional $P\to Q$?

Remember, the only time the conditional is false is if it is cloudy and it does not rain, that is, if $P \land \neg Q$. However, if we have $P$ or $\neg Q$ we could have that situation if both of these are true. Then the conditional would be false. In that case $P \lor \neg Q$ cannot represent the conditional $P\to Q$.

Answer 4

There are often multiple conceptualizations that can lead to different intuitive understandings. This is the constructive (intuitionistic) one. Constructively speaking:

$$P \Rightarrow Q$$

means "If I can build a $P$ then I can build a $Q$", or equivalently, "I can build a machine that inputs a $P$ and outputs a $Q$".

$$\lnot P \lor Q$$

means "I can't build a $P$, or I can build a $Q$" (see caveat below). Equivalent to the above: if you can't build a $P$, then any device serves as an example: you'll never be able to prove that it doesn't output a $Q$ when given a $P$, since you can't produce the $P$. On the other hand, if you can build the $Q$, then you can also build a device which inputs a $P$ and outputs a $Q$: just ignore the input.

$$P \lor \lnot Q$$

means "I can build a $P$, or I can't build a $Q$" (same caveat below). Just because you can build a $P$ doesn't mean that $P$ let's you build a $Q$. And if you can't build a $Q$ (that is $\lnot Q$), then you can't build it even if you can build a $P$.

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Caveat: $A \lor B$ actually has to mean "I can build an $A$ or I can build a $B$ and I know which one I can build" for it to be a constructive assumption. That is why $P \implies Q$ and $\lnot P \lor Q$ are not exactly equivalent constructively: the second implies the first, but the first doesn't tell you whether $\lnot P$ or $Q$ is the one that is true.

On the other hand, $\{P \lor \lnot P, P \implies Q\}$ is constructively equivalent to $\lnot P \lor Q$ because the knowledge of which of $P \lor \lnot P$ is true is enough to know which of $\lnot P \lor Q$ is true.