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So I am told by a friend that "axioms in an axiomatic system cannot be proved within the axiomatic system". I was wondering how true this is. Is there any actual mathematical theorem that states something like this.


EDIT: Along the same lines, how true is it to say that "an axiom in an axiomatic system cannot be disproved within the axiomatic system"?

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"axioms in an axiomatic system cannot be proved within the axiomatic system".

See Aristotle, Post.An, Bk.I, 82a7-82a9:

This is the same as to inquire whether demonstrations go on ad infinitum and whether there is demonstration of everything, or whether some terms are bounded by one another.

There are two uses of "proof" here: the usual one and the formal one.

In a formal system a proof is derivation in the system, i.e. a sequences of formulas where every formula either is an axiom or is derived from previous ones in the sequence by way of rules of inference.

The conclusion of the proof, i.e. the last formula in the sequence, is a theorem.

Thus, formally speaking, a one-line derivation, where necessary the formula is an axiom, is a formal proof whose conclusion, the axiom itself, is a theorem of the system.

But obviously in the common sense meaning of "proof", the above derivation will not be considered an "interesting" proof of the axioms:

A mathematical proof is an inferential argument for a mathematical statement, showing that the stated assumptions logically guarantee the conclusion. The argument may use other previously established statements, such as theorems; but every proof can, in principle, be constructed using only certain basic or original assumptions known as axioms, along with the accepted rules of inference.

Proofs are examples of exhaustive deductive reasoning which establish logical certainty.

  • Okay interesting... So what I understand is that an axiom is a one-line proof of itself, is that what you meant? Cool... so because an axiom is a proof of itself can we think of that as a reason why axiom is always true? – TheImperfectCrazy Mar 31 '21 at 16:55
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Perhaps the more interesting question is whether some axiom can be proved using the other axioms, without that axiom in the system. A proof system can prove any of the axioms, since they are axioms.

And it is not a trivial question, a very long time went into trying to prove the parallel postulate from the other axioms of Euclidean geometry. Later a sound proof was provided that this could not be done.

Today we have better theory to know which axioms are redundant and which are not. Usually upon building an axiom system the redundant axioms are removed. However, this is not always the case, for example even in set theory the axiom of empty set can be either proved or left as an axiom.

Furthermore, we can't really prove that number theoretic systems are consistent without assuming even more axioms that can't be proven consistent. So ultimately it is possible that some of the axioms are still redundant and even cause inconsistency, even if almost no one intuitively believes this.

Dole
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  • I didnt quite understand the first paragraph. Could you please elaborate that a bit more. "some axiom can be proved using the other axioms, without that axiom in the system" - What I understand of this is - we have say one axiom each in two different axiomatic system and we are trying to prove one axiom in a system using a different axiom from a different system. Is that what you mean? – TheImperfectCrazy Mar 31 '21 at 17:06
  • @TheImperfectCrazy Yes, so if we have axioms $A_1, A_2, A_3$, and now I remove $A_2$, can we prove $A_2$ using $A_1$ and $A_3$, e.g. is $A_2$ independent. – Dole Mar 31 '21 at 17:18
  • @TheImperfectCrazy It is also possible to "prove" axioms that denote the same thing in a other system. For example, in set theory you prove the induction principle, while in $PA$ it is an axiom. But anyway, I think your friend is just saying that the axioms are taken on faith or something to that effect. – Dole Mar 31 '21 at 17:23
  • Ahh okky great. I understood. Yes so you are right, I agreed to what he said in that context, but then it made me think how vigorous is that statement. Hey also, I just now thought of a similar question and I have edited the OP to include that as well, could you please comment on "an axiom in an axiomatic system cannot be disproved within the axiomatic system" if you have time. – TheImperfectCrazy Mar 31 '21 at 17:40
  • @TheImperfectCrazy I will just comment here, axiom can be disproved in an axiom system, in fact even by a other axiom. Then the axiom system is inconsistent and in general can prove anything. (Disproved in the sense of proving the negation). – Dole Mar 31 '21 at 17:44