Logic is used to make arguments every day, and typically we don’t need to think too hard about the arguments we are making. Who has ever thought hard to analyze systematically the statement, “if I want to make tea, then I must have hot water?” In mathematics and physics, however, we need to be much more careful with the arguments we make. So, it’s worth taking the time to think about how to abstract and formalize the arguments we make.

Students in mathematics are typically taught systematic logic as a pre-requisite to any advanced coursework beyond, say, calculus. In physics, however, abstract logic isn’t typically given much emphasis. This doesn’t present much of an issue most of the time since the emphasis in physics is placed on calculation rather than proof, at least at the introductory level. But even in physics formal results require proof to be accepted, and that means argument needs to be understood eventually.

All arguments consist of combinations of statements. Let’s say \(P\) is some statement. For example, we could have \(P\equiv\)”the sky is blue” and we could then ask whether the statement \(P\) is true of false. To stick to more math, we can think about statements more like \(P\equiv”\exists x\in\mathbb{R}, x^2+1=3\). That is, the statement \(P\) is “there exists a real number \(x\) such that \(x^2+1=3\)”. Now, we know this \(P\) happens to be true since both \(\pm\sqrt{2}\) are real numbers, but we are still allowed to ask the question “is \(P\) true” and perhaps in some more complicated situation, we might not be sure.

It will be helpful to have an example as we go through this, and in fact this is the example which made me write the original version of this post. We are going to think about the Fourier transform:
$$\phi(\omega)=\int_{-\infty}^\infty\psi(x)e^{-ipx}\frac{\text{d}x}{\sqrt{2\pi}}$$.
Our goal will be to prove that
$$\psi(x)=\int_{-\infty}^\infty\phi(p)e^{ipx}\frac{\text{d}p}{\sqrt{2\pi}}$$.
Essentially, we are to show that if \(\phi\) is the Fourier transform of \(\psi\), then the Fourier transform may be inverted by the second expression.

Putting this into the language of predicate logic, we could take both of the equalities above to be statements:
$$P\equiv”\phi(\omega)=\int_{-\infty}^\infty\psi(x)e^{-ipx}\frac{\text{d}x}{\sqrt{2\pi}}”$$,
$$Q\equiv”\psi(x)=\int_{-\infty}^\infty\phi(p)e^{ipx}\frac{\text{d}p}{\sqrt{2\pi}}”$$.

Now, we note that either of these statements could be true or false depending on what $\psi$ and $\phi$ are. For example, if \(\phi(p)=1\) and \(\psi(x)=1\), \(P\) would immediately be false (as, incidentally, would \(Q\)). With this, the thing we are asked to prove may be written \(P\Rightarrow Q\) independent of the truth of P and Q. That is, we want to be able to rely on the statement \(P\Rightarrow Q\) without needing to check \(P\) or \(Q\).

So then, we need to know the truths of \(P\) and \(Q\) influence the truth of the statement \(P\Rightarrow Q\). This information can be summarized in a truth table:

\(P\)	\(Q\)	\(P\Rightarrow Q\)
T	T	T
T	F	F
F	T	T
F	F	T

From this, we can note that if \(P\) is false, it already doesn’t matter what the truth of \(Q\) is, the implication will be true. As a result, we only have to check what happens when \(P\) is true. In particular, if by assuming \(P\) to be true we are able to show that \(Q\) must also be true (without making any assumptions about \(Q\)), we would have precluded the only possible case in which the implication could have turned out to be false.

One way to think about this is to consider the truth table above to give the truth values of the implication if the truths of \(P\) and \(Q\) were independent of each other. What we want to show is, in essence, that the truths of \(P\) and \(Q\) are not independent, but instead conspire to be related in such a way that the implication is never false.

Now, since we are allowing ourselves, based on our prior considerations, to assume \(P\) to be true, in the case of the Fourier transform, we are allowed to assume that \(\phi\) and \(\psi\) are related to each other by
$$\phi(\omega)=\int_{-\infty}^\infty\psi(x)e^{-ipx}\frac{\text{d}x}{\sqrt{2\pi}}$$.

Here is where the potentially subtle point comes in: while we cannot assume that \(Q\) is true, there is absolutely no issue thinking about the quantity \(\int_{-\infty}^\infty\phi(p)e^{ipx}\frac{\text{d}p}{\sqrt{2\pi}}\). The only issue would be if we tried to assert that this equaled \(\phi(x)\), which we cannot do because that would be equivalent to the assertion that \(Q\) is true. We want to show that \(Q\) is true, so it would be singularly pointless to assume it to be true.

This is a sticking point I think many students who are new to logic struggle with. Simply because we are to assume \(P\) to be true doesn’t mean we need to start with it as our first line of work, we are just allowing ourselves the ability to use it in some way at some point. In fact, for this proof, we will use the truth of \(P\) in the very next line to replace \(\phi\) by its Fourier integral:

$$\int_{-\infty}^\infty\phi(p)e^{ipx}\frac{\text{d}p}{\sqrt{2\pi}}=\int_{-\infty}^\infty\left[\int_{-\infty}^\infty\psi(x)e^{-ipx}\frac{\text{d}x}{\sqrt{2\pi}}\right]e^{ipx}\frac{\text{d}p}{\sqrt{2\pi}}=\int_{-\infty}^\infty\left[\frac{1}{2\pi}\int_{-\infty}^\infty e^{ip(x-x^\prime)}\text{d}p\right]\psi(x^\prime)\text{d}x^\prime$$.

But, if we now recall the Fourier representation for the Dirac delta function,
$$\delta(x-x^\prime)=\int_{-\infty}^\infty e^{ip(x-x^\prime)}\text{d}p$$,
we may replace the bracketed expression by the Dirac delta to find
$$\int_{-\infty}^\infty\phi(p)e^{ipx}\text{d}p=\int_{-\infty}^\infty\delta(x-x^\prime)\psi(x^\prime)\text{d}x^\prime=\phi(x)$$
which establishes that, indeed, \(Q\) is a true statement.

To review what we have done here, we assumed the truth of \(P\) because when \(P\) is false the implication is always true anyway. Given this assumption, we showed that \(Q\) is also always true, precluding the only possible way for an implication to be false.

The argument we have given very carefully avoids making any assumptions about \(Q\) and it should also be pointed out that just showing \(P\Rightarrow Q\) to be true tells us nothing about the truth of \(Q\Rightarrow P\) (though in this particular example, it happens to also be true and can be shown by an analogous argument).