On pp. 26-27 of his Introduction to Mathematical Logic (5th edition), Elliott Mendelson writes:
- If $\mathscr{B}$, $\mathscr{C}$, and $\mathscr{D}$ are wfs of $\mathrm{L}$, then the following are axioms of $\mathrm{L}$:
(A1) $(\mathscr{B} \Rightarrow (\mathscr{C} \Rightarrow \mathscr{B}))$
(A2) $((\mathscr{B} \Rightarrow (\mathscr{C} \Rightarrow \mathscr{D})) \Rightarrow ((\mathscr{B}\Rightarrow\mathscr{C})\Rightarrow(\mathscr{B}\Rightarrow\mathscr{D})))$
(A3) $(((\neg\mathscr{C})\Rightarrow(\neg\mathscr{B}))\Rightarrow(((\neg\mathscr{C})\Rightarrow\mathscr{B})\Rightarrow \mathscr{C}))$
In my prior experience, axioms have always been chosen so as to closely mirror our intuitions. E.g. "things that are equal to the same thing are also equal to each other."
In contrast, I find the axioms quoted above completely opaque.
I fail to see the purpose of grounding a theory on such opaque statements. Sure, one can derive "results" from them, but who cares? In other words, what makes the results derived from axioms A1, A2, and A3 above more worthy of our interest than results derived from some other random set of equally obscure axioms?
EDIT: Thank you for all the answers and comments! They have indeed been extremely instructive. I now have a better idea of what is tripping me up. It all boils down to the conflict I perceive between the word "axiom", as I have come to understand it from my prior experience (e.g. with Euclid's axioms or Peano's axioms), and Mendelson's A1, A2, and A3.
More specifically, the "axioms" that Mendelson proposes lack two "defining features" of my prior understanding of the word "axiom."
The first of these is that axioms are supposed to codify "plain common sense." No special training is required to accept them as true.
The second defining feature of my prior idea of an "axiom" is that it cannot be justified by appeal to even more obviously true facts. An "axiom" is not only obviously true, but also irreducibly so.
After reading your comments and answers, I think that I can come up with a reasonable explanation for the first discrepancy. It goes something like this: in this field the word "axiom" is chosen only (or primarily) for its connotation as "starting point". All other traditional connotations of the word "axiom" (such as "being commonsensical") are left out. In this new usage, "axiom" is, basically, a "term of art". It is suggestive, in the same way that words like "expression," "clause," and "statement" are when used to describe a computer "language". The meanings of all these technical words have only a distant resemblance to those of their natural language namesakes.
I have a harder time disposing of the second issue. What still confuses me is that, in discussions of these axioms, I often come across appeals to using truth tables as a way to convince oneself that these axioms are true. This suggests to me that these "axioms" may be true, but they are certainly not fundamental, since we are justifying them by appeal to something else, namely whatever we appeal to when we use truth tables. More specifically, it appears as though the real axioms here (in the sense of "rock-bottom principles") are the truth tables for $\neg$ and $\Rightarrow$.
Best Answer
I always think of the first axiom as a kind of Conditionalization:
$P$
$\therefore Q \to P$
Conditionalization allows you, in effect, to bring results inside a certain context. That is, once we know that $P$ is true, then within the context of $Q$, we still have $P$.
Now, instead of as an inference, we write this as a single statement $P \to (Q \to P)$, but note that together with the Modus Ponens (MP) inference rule that almost all these kinds of axiom systems have, you can of course now make exactly the above inference:
$P$
$P \to (Q \to P)$ (axiom)
$\therefore Q \to P$ (MP)
The second axiom is a kind of Conditionalized Modus Ponens. In effect, it points out that we can still do Modus Ponens inside a certain context ($P$). That is, the Conditionalized version of Modus Ponens:
$Q \to R$
$Q$
$\therefore R$
becomes:
$P \to (Q \to R)$
$P \to Q$
$\therefore P \to R$
And again, we can capture this with a single statement: $((P \to (Q \to R) \land (P \to Q)) \to (P \to R)$ which by Exportation is equivalent to:
$((P \to (Q \to R) \to ((P \to Q) \to (P \to R))$
(and you really want the latter, since you want to express everything with $\to$'s and $\neg$'s)
Because of these two properties, we can prove the Deduction Theorem as a result of these two axioms:
If $\Gamma, \phi \vdash \psi$, then $\Gamma \vdash \phi \to \psi$
That is: If we can derive $\psi$ from $\Gamma$ within the further context of $\phi$, then we can derive the conditionalized version of $\psi$, i.e. $\phi \to \psi$ from $\Gamma$ alone.
This important meta-logical result is why you so often see these first two axioms included in the various axiom systems.
Now, the third axiom is, by itself, probably the most understandable one: it expresses the Reductio Ad Absurdum: we can prove $P$, if we can show that its opposite $\neg P$ leads to a contradiction (i.e. if $\neg P$ leads to both $Q$ and $\neg Q$ for some statement $Q$:
$\neg P \to Q$
$\neg P \to \neg Q$
$\therefore P$
And once again, we can capture this with a single statement (reversing the two premises): $((\neg P \to \neg Q) \land (\neg P \to Q)) \to P$ which (again by Exportation) is equivalent to:
$(\neg P \to \neg Q) \to ((\neg P \to Q) \to P)$
Now, what's cool about this is that together, these three axioms actually become a complete system, as I am sure other Answers will point out. But I thought I would try to provide you with a more conceptual understanding of the axioms themselves, which is I believe you were really asking for.