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Although impractical, propositional logic can be formulated with purely the set of logical connectives {$\neg,\rightarrow$} (or even just NAND/NOR), along with a sufficiently large set of propositional atoms, a set of logical axioms commonly stated as:

  • $\vdash p\rightarrow(q\rightarrow p)$
  • $\vdash(p\rightarrow(q\rightarrow r))\rightarrow((p\rightarrow q)\rightarrow(p\rightarrow r))$
  • $\vdash(\neg p\rightarrow\neg q)\rightarrow(q\rightarrow p)$

And a set of rules of inference ($\textit{modus ponens}$):

  • $p,\space p\rightarrow q\vdash q$

Is this sufficient to characterize all of propositional logic? In other words, can every proof in propositional logic be reduced to combinations of solely these symbols and applications of purely these rules? I'm aware every formula can be reduced to combinations of $\neg$ and $\rightarrow$, or to define other logical connectives in terms of only these for ease of use, for example:

  • $p\land q:=\neg(p\rightarrow\neg q)$
  • $p\lor q:=\neg p\rightarrow q$

Wikipedia gives a list of common rules of inference here, is it possible to do a similar thing and define all of these rules of inference using only the information given in the introduction, i.e. take, for instance, disjunctive syllogism:

$$\frac{p\lor q,\space\neg p}{\therefore q}$$

Is it possible to prove

$$\frac{\neg p\rightarrow q,\space\neg p}{\therefore q}$$

And any other of the given rules on that page? If not, what is the minimal set of rules of inference? I ask this as a mental exercise, obviously it is not practical to reduce everything to this smallest sufficient set of rules, in the same way that we use $\land$, $\lor$, and $\leftrightarrow$ in addition to the other two for ease of use.

Also, if my understanding is right, we can take any two of the three axioms and turn them into rules of inference as long as we leave at least one element in our set of axioms, but please correct me if I'm wrong.

  • may not apply here, like it did to a question about logical systems without modus ponens earlier but :https://plato.stanford.edu/entries/logic-paraconsistent/ –  Jul 30 '17 at 02:20
  • @RoddyMacPhee Interesting, but I am speaking within the framework of classical logic. –  Jul 30 '17 at 03:07

2 Answers2

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Correct! Those 3 axioms plus modus ponens form a complete proof system: every valid argument involving $\neg$ and $\rightarrow$ can be proven to be valid using just those 3 axioms plus modus ponens. And since $\{ \neg, \rightarrow \}$ is an expressively complete set of operators, any propositional logic operator can be defined in terms of them, and hence any argument as well and again: if it is valid, it can be proven to be valid within this very simple system (the proofs themselves, of course, are often far from simple! Though there are algorithms to automatically create a proof for any valid argument)

You also asked about transforming axioms into inference rules. I am not exactly sure what you mean, but obviously each axiom can be treated as an inference rule: from nothing, you can infer a statement of the form as indicated.

Oh, and please note: the letters $p$, $q$, and $r$ are sentence variables, meaning that you can fill in any sentence for them. Thus, for example:

$$(A \rightarrow \neg B) \rightarrow (\neg \neg C \rightarrow (A \rightarrow \neg B))$$

would be an instance of axiom 1, where we use $A \rightarrow \neg B$ for $p$ and $\neg \neg C$ for $q$

Bram28
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Yes, it is possible. These axioms and modus ponens are suficient. Remember that you have Deduction theorem and these rules like disjunctive syllogism can be turn axioms. Disjunctive syllogism is as example $(p \lor q) \to (\neg p \to q)$

Thiago Nascimento
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  • Another example $p \to (p \lor q)$ = $p \to (\neg p \to q)$. Using explosion, this is a theorem. To prove explosion, see that $(p \to p)$ is a theorem and double-negation too. Now, using $\neg (p \land \neg p) = \neg \neg (p \to \neg \neg p) = p \to p$, therefore $\neg (p \land \neg p)$ is a theorem. Explosion: $ (\neg q \to \neg( p \land \neg p)) \to (( p \land \neg p) \to q)$, using modus ponens you have $(p \land \neg p) \to q$ – Thiago Nascimento Jul 30 '17 at 03:14