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Suppose I have two mathematical statements : $A$ and $B$.

Suppose that $A$ is an already proven theorem.

Suppose that, to prove $B$, I use $A$ somewhere in the proof.

Therefore, $B$ is a proven theorem itself.

Now, suppose I want to find another proof for $A$. Specifically, I prove $A$ by resorting to $B$ somewhere in the proof.

Can I say that I have produced another valid proof for $A$ ?

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I wouldn't say so... Why do you need to know, anyway? –  user148697 May 14 at 13:34
    
@jonnytan999 I need to know, because I'm facing this exact issue right now... –  G.T.R May 14 at 13:36
    
What do you mean? Are you asked to find multiple proofs for some statement? Also consider the case $B = A \wedge (1 = 1)$ (or something to that effect). Wouldn't that feel weird to say you have a "new" proof in this case? –  Najib Idrissi May 14 at 13:49
    
@NajibIdrissi I was wondering if there were a simple proof for Stone Weierstrass theorem (see here math.stackexchange.com/questions/755130/… ). Stone Weierstrass theorem($A$) itself proves the existence of points that grant uniform convergence of the sequence of Lagrange polynomials ($B$). $B$, in turn, grants a proof of Stone Weierstrass ... –  G.T.R May 14 at 13:57
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When you prove a theorem A, you should be able to build a tree. You list all of the theorems, and axioms used to prove it. Then for each of the theorems you list all of the theorems and axioms in that, etc. You continue this until you're left with just axioms. If anywhere in that tree, you arrive at A, then your tree is "circular" and actually never ends. –  Cruncher May 14 at 15:46

3 Answers 3

up vote 3 down vote accepted

The name for this fallacy is begging the question/petitio principii.

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In what way ? What I fail to understand is : once $B$ is proven, isn't $B$ true by itself independently of $A$ ? –  G.T.R May 14 at 13:44
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$A$ and $B$ are theorems, yes. But your proof of $A$ isn't valid. If you prove $A\iff B$, then $A$ and $B$ will be true or false simultaneously. –  Martín-Blas Pérez Pinilla May 14 at 13:51
    
This isn't correct. –  Doug Spoonwood May 14 at 14:26
    
@Martín-BlasPérezPinilla That A and B end up true or false simultaneously doesn't just happen because of an equivalence. In the pure implicational calculus of propositions all theorems are simultaneously true, even though no equivalence lies around. –  Doug Spoonwood May 14 at 14:42
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@DougSpoonwood, the pure implicational calculus of propositions has some relevance for this case? –  Martín-Blas Pérez Pinilla May 14 at 15:05

Short answer: Not really.

Long answer: Not really. If you want to be pedant, it's a "new proof": you prove that $A$ is true, then you prove something else, and then you conclude that $A$ is true... instead of concluding that $A$ is true directly. It's as if you took a proof, inserted somewhere "therefore $1=1$", and then went on. It's a "new" proof, from a formal point of view, but any mathematician would (rightly) say it's not.

In other words, you have some set of axioms $T$. You can find deductions in a formal system that $T \vdash A$, $T \cup \{A\} \vdash B$ and $T \cup \{B\} \vdash A$. Then you can put them back to back (using modus ponens) to deduce something like $T \vdash ((A \wedge (A \implies B) \wedge (B \implies A)) \implies A)$. But in doing so, you're simply proving $A \implies A$.

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Let's say our only axiom is A: (p$\rightarrow$(q$\rightarrow$(r$\rightarrow$q))) and we have uniform substitution and detachment as our only rules of inference. We can obtain B: (r$\rightarrow$(s$\rightarrow$r)) as follows:

axiom 1           1 (p→(q→(r→q)))
by axiom 1        2 ((p→(q→(r→q)))→(r→(s→r)))
detachment 2 ,1   3 (r→(s→r))

To prove B (r$\rightarrow$(s$\rightarrow$r)) we used A (p→(q→(r→q))) in the proof. Now let's prove A by restoring to B somewhere in the proof as follows.

theorem             1 (r→(s→r))
1 r/(r→(s→r)), s/q  2 ((r→(s→r))→(q→(r→(s→r))))
detachment 2, 1     3 (q→(r→(s→r)))
3 q/p, r/q, s/r     4 (p→(q→(r→q)))

In this case you have produced another valid proof for A (the original proof of A consisted of listing it as an axiom). So, the answer to your question is "yes" in certain systems of logic.

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It seems part of your answer has disappeared. –  G.T.R May 14 at 14:31
    
@GabrielR. I intentionally put up half an answer for a moment so I could copy and paste instead of having to deal with the TeX code. Thanks though! –  Doug Spoonwood May 14 at 14:40
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But you used $A$ to prove $B$, so you get no independent proof of $A$. You have established the equivalence of $A$ and $B$ on the basis of the other axioms (none) and rules of inference. –  Daniel Fischer May 14 at 14:47
    
@DanielFischer The assumption of the question comes as that A gets used to prove B. Now, the second proof only needs B in order to prove A. Consequently, those proofs do qualify as independent (though the two formulas are not). Also, there is NO such thing as equivalence in this logic (and you can't represent logical equivalence using just implication... at least in some logical systems), there only exists implication. Consequently, I have NOT established an equivalence of A and B (though they both hold as theorems simultaneously). –  Doug Spoonwood May 14 at 14:52

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