If/Then Statements (P⇒Q); Transitivity

An implication is a relation between $2$ propositions. Suppose that we have $2$ propositions $P$ and $Q$, and we wish to express this relation. This can be stated in a number of ways. The forms of the implication that are in bold are the ones most commonly used:

$P$ implies $Q$.

$Q$ is implied by $P$.

If $P$, then $Q$.

$Q$, if $P$.

$P$ only if $Q$.

When $P$, $Q$.

There are still many more ways, but these are the simplest ones, and the two that are in bold are used most frequently by far. In the above examples, $P$ is referred to as the antecedent, and $Q$ is referred to as the consequent.

Symbolically, the implication is written as $P\implies Q$ or $P\supset Q$. The first one is far more common, and it better evokes the meaning of an implication. In addition, the second one can easily be confused with "$P$ is a superset of $Q$," which is basically the opposite of what we want, so I will use only the first one from now on.

An implication $P\implies Q$ intuitively means that $P$ being true "leads to" $Q$ being true. It's actually quite difficult to define an implication in this manner, because there need not be any meaningful relationship between the actual meanings of the statements $P$ and $Q$ in written language.

Thus, we will define an implication using its truth value, which will be based solely on the truth values of its component statements.

This might have been a little bit tricky! It turns out that the implication is defined to be false when $P$ is true and $Q$ is false and true in all other cases!

Why is this? Well, if $P$ is true, then since $P$ "leads to" $Q$, then $Q$ should be forced true as well! Thus, when $P$ is true and $Q$ is false, it means that $P$ cannot imply $Q$.

However, in all of the other cases, we cannot prove that $P$ does not imply $Q$. The truth is that we can never prove rigorously that one thing causes another. Thus, if the truth values do not automatically exclude the implication from existing, then we say that the implication is true. We will show this by thinking of the implication as an "obligation" that needs to be fulfilled if the conditions are met (perhaps "If I give you a gift, you need to give me a gift in return" or something similar).

First, suppose that the consequent is true. This would mean that the obligation has already been fulfilled, regardless of whether the conditions were even satisfied! There is no way to observe whether $P$ leads to $Q$ if $Q$ is going to occur anyway! This is why we consider $P\implies Q$ true if $Q$ is true, regardless of the truth value of $P$. This can be easily remembered by using the notion: a true statement can be implied by anything.

Now, suppose that the antecedent is false. This would mean that the conditions have not even been met. Thus, who cares if we fulfill our obligations? There is no way to observe whether $P$ leads to $Q$ if we can't even observe $P$ occur! This is why we consider $P\implies Q$ true if $P$ is false, regardless of the truth value of $Q$. This can be easily remembered by using the notion: a false statement can imply anything.

Note that $P\implies Q$ is false in the case where $P$ is true and $Q$ is false, and it is true otherwise. This is expressed in the Law of Material Implication, which states the following: $$P\implies Q\equiv \neg P\vee Q$$ This equivalence describes the meaning of an implication in terms of truth value, and it is called this for that reason. In fact, the implication in general in this context is often called a material implication, to distinguish it from logical consequences and causal relationships, since it only depends on the truth values and not on the relationship between the statements.

An important rule that implications satisfy is the Law of Transitivity, also called the Hypothetical Syllogism, which states the following: $$P\implies Q$$ $$Q\implies R$$ $$\therefore P\implies R$$

This intuitively makes sense: if $P$ being true will force $Q$ to be true, and $Q$ being true will force $R$ to be true, then it seems reasonable that $P$ being true will force $R$ to be true, by "passing through" $Q$.

This can be proved by using the truth values of the statements. $P\implies R$ is true if $P$, $Q$, and $R$ are, respectively, $TTT$, $TFT$, $FTT$, $FFT$, $FTF$, $FFF$ ($T$=true and $F$=false), and it is false in all other cases. Now, look at the statement $(P\implies Q)\wedge(Q\implies R)$. This statement is true if $P$, $Q$, and $R$ are, respectively, $TTT$, $FTT$, $FFT$, $FFF$ ($T$=true and $F$=false), and it is false in all other cases. In all of the cases where $(P\implies Q)\wedge(Q\implies R)$ is true, $P\implies R$ is true as well. Therefore, $$P\implies Q$$ $$Q\implies R$$ $$\therefore P\implies R$$ There are many ways of proving transitivity of implications, most of which are more or less the same. This is just one of them.