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Let $A$ be a statement for which is not a tautology. Let $L^+$ be the formal theory obtained from $L$ by adding as new axioms all formulas obtainable from $A$ by substituting arbitrary statement forms for the statement letters in $A$, the same forms being substituted for all occurrences of a statement letter. Show that $L^+$ is inconsistent.

I have several questions:

  1. can $A$ be an axiom by itself from given conditions ?
  2. consider given assignment of truth values to the statement letters that gives us $A = false$. Let's replace statement letters of $false$ value with negation of a tautology, and of $true$ values with a tautology. So how can this obtained formula be an axiom that is be of $true$ value?

I cannot understand how to show that $L^+$ is inconsistent.

user142248
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  • What is a statement letter? Is it a propositional variable like $P$ in $P \to Q$ or a metavariable like $\varphi$ in $\varphi \land \lnot \varphi$ or something different? – Greg Nisbet Jul 13 '21 at 15:28
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    yes, they are as I got it from "Introduction to mathematical logic" by Mendelson, Elliott – user142248 Jul 13 '21 at 15:35
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    btw do we need to consider that axioms must be of true value when it comes to inference of formulas ? – user142248 Jul 13 '21 at 15:42
  • The question is tagged first-order-logic, but, on further reflection, I'm not sure that's right. Is the question about first-order logic or is it about different propositional logics? Can you give us some examples of possible $L$s? – Greg Nisbet Jul 13 '21 at 15:49
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    Sorry, English isn't my tongue language, it was misleading. It's about propositional-calculus – user142248 Jul 13 '21 at 16:09
  • Thank you. Also, on this site, the consensus is that questions can be asked in any language. Someone, either the author or a community member, can then add a translation into English at the end of the post. For questions, you have a choice between using English and using other languages. For answers, I don't know what the consensus is. – Greg Nisbet Jul 13 '21 at 16:17

1 Answers1

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can A be an axiom by itself from given conditions ?

Yes, it will be an axiom -- because any "statement letter" is in itself a "statement form", and substituting them all with themselves is a perfectly good invocation of the procedure in the definition.

The second part of your question is in fact the solution:

consider given assignment of truth values to the statement letters that gives us $A = false$. Let's replace statement letters of $false$ value with negation of a tautology, and of $true$ values with a tautology. So how can this obtained formula be an axiom that is be of $true$ value?

Remember that "axiom" doesn't mean "necessarily true" -- it simply means "something we allow a proof to start from without further argument".

You have the right idea that $L^+$ has an axiom whose truth value is false under all truth assignments.

Depending on how you define "inconsistent", you're either done or almost done. Namely, if "inconsistent" mean that the theory has no satisfying truth assignments, then you have argued for exactly that.

On the other hand, if (as is more common) "inconsistent" means that you can derive a contradiction, then note that your variant of $A$ is always false, and therefore the negation of that formula is always true, that is, a tautology. If your proof system is complete (which I will assume you know it is), that means that you can derive that tautology. The axiom itself, by virtue of being an axiom, can of course also be derived. In total, you have derived a both a formula and its negation, and thereby proved that $L^+$ is inconsistent.

Troposphere
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  • ok, according to the conditions it's possible to construct an axiom $A$ which is always $false$ and as it is the axiom it can be derived. On the other hand I can take $\neg A$ which is tautology and as $L^+$ includes $L$ that is $L^+$ is complete it means that $\neg A$ is derivable from $L^+$. So both $A$ and $\neg A$ can be derived that is $L^+$ is inconsistent. – user142248 Jul 14 '21 at 22:41