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In hmakholm's answer to this question, the following is written:

If it has no axioms then there is no way to begin a proof in the empty theory, and without rules of inference all that could be proven would be the axioms themselves.

I don't see how either of these assertions can be true. Tackling the first; let's say I have a deductive system which has the following rule, among other:

$$\Gamma \vdash \theta \ \text{and} \ \Gamma \vdash \phi, \ \text{then} \ \Gamma \vdash \theta \land \phi $$

Well, whenever I wind up with the two premises, I am able to prove $\theta \land \phi$, right?

Now, to tackle the second; let's say my logic has the following axiom:

$$P \lor \neg P$$

Let's say I find out, or assume, that $n$ is even. Well, now I have to premises; an axiom and a fact/assumption.

$$P \lor \neg P \\ 2 \mid n \\ \therefore \neg(2 \not \mid n)$$

Surely, $\neg(2 \not \mid n)$ is not an axiom? It follows from premises, unlike axioms, so it cannot be an axiom, right?

user110391
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    In your first example, you have a rule and in your second example you have an axiom. How do these contradict the statement that you need rules and axioms? – John Douma Sep 24 '22 at 06:03
  • @JohnDouma I think the OP is suggesting that in the first example there is no axiom and in the second, no rule of inference. – David Sep 24 '22 at 06:10

2 Answers2

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(1) Where did $\Gamma$ come from? Ultimately it must have come from an axiom (if it is not an axiom itself).

(2) You are using a rule of inference: from $\Phi$ deduce $\neg(\neg\Phi)$.

And BTW, "$n$ is even" is expressed by $2\mid n$, not $n\mid2$.

David
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  • This is what I suspected. I am just having trouble with this since they are fundamentally the same thing; assumptions. But I guess, if I am to deduce something from an axiom, I must have a rule for how to convert that axiom into an equivalent proposition. – user110391 Sep 24 '22 at 06:14
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Natural Deduction systems like Fitch-Style systems have no axioms, but plenty of inference rules. However, some proofs in such a system might not be regarded as such to some critics. Take the two following proofs:

  1. $p$ (Hypothesis)
  2. $p \to p$ (1-1 $\to$ Intro)

and

  1. $(p \land \neg p)$ (Hypothesis)
  2. $\neg (p \land \neg p)$ (1-1 $\neg$ intro)

These are valid proofs in such a calculus, and require no axioms. I do think they are a bit convenient, are not very informative, and are still presuppositions that go into ND.

PW_246
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