Definition – Mathematical implication.

Definition

A mathematical implication is a statement of the form $p\implies{q}$ . The truth values of an implication are given by

$\begin{array}{c|c|c} p & q & p\implies{q} \\\hline T & T & T \\ T & F & F \\ F & T & T \\ F & F & T \end{array}$

Where $p$ and $q$ are mathematical statements. We refer to the statement $p$ as the antecedent, and we refer to $q$ as the consequent. Finally, we say that $p$ implies $q$ .

Notes

Many courses and books simply use the truth table above to define the implication. This does not, however, give any insight into why an implication is defined the way it is. What’s more, it definitely does not go into anywhere near enough depth to explain “vacuous truths” [definition coming soon]. An attempt will be made here to clarify this so called definition.

Imagine that we deduce a statement $q$ from a statement $p$ . If this deduction is valid and logical, it makes sense to seek some notation to denote this. Here comes the $\implies$ symbol. If $p$ is true and $q$ is true, it makes sense that our deduction is true, and so we define $p\implies{q}$ to be true. Hence the first row.

Secondly, if a statement $p$ is true, and using some deduction we end up at a false statement, then our deduction must have been false. This is true because we could not get from $p$ to $q$ using valid logic. It therefore makes sense for the statement $p\implies{q}$ to be false, to denote the fact that our logical deduction was invalid.

Now, imagine that we go from a false statement to a true statement. If we apply valid logic to a false statement, it is still possible to arrive at a valid statement $q$ . For example, consider the statement $5 = 7$ . One axiom of the field of real numbers is that for every element, say $x$ , of the field’s set $A$ , there exists a corresponding element $-x\in{A}$ such that $x+(-x)=0$ . Now, since $5=7$ , it follows that they are the same element of the field’s set (by definition of equality), $A$ , and therefore $5=7=x\in{A}$ . It follows that there exists an element $-5=-7=-x\in{A}$ such that $5+(-5)=7+(-7)= 0$ . Ie, the implication $5=7\implies{0=0}$ is true. This is an explanation of the third line.

Similarly, if our statement $p$ is false and our implication is true (because we used a valid deduction), then our statement will still be false. This is an explanation for the fourth line of the definition, and completes the understanding. We write $p\implies{q}$ if the deduction used to get from $p$ to $q$ is logically valid.