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The problem goes like this: "Assume $A$$1$ $\equiv$ $A$$2$ . Show that for any formula $C$-containing $A$$1$ as a part , if we replace one of more occurences of the part $A$$1$ by $A$$2$ , then the resulting formula is logically equivalent to $C$."

Now, at first glance ,this theorem seems obvious to me , but I still have to prove it. My choice is to use the law of induction and the following identities (which can be derived from $A$$1$ $\equiv$ $A$$2$).

$1$. $A$$1$ $\equiv$ $A$$2$
$2$. ($\lnot$$A$$1$) $\equiv$ ($\lnot$$A$$2$)
$3$. ($A$$1$) $d$ $B$ $\equiv$ ($A$$2$) $d$ $B$
$4$. $B$ $d$ ($A$$1$) $\equiv$ $B$ $d$ ($A$$1$)
(Here $d\in\{\land,\lor,\Leftarrow,\Rightarrow,\Leftrightarrow\}$)

Now, I am used to doing induction with only one operation (like $+$), but now there is more then one operation. Also, I have to to consider commuting (rule $3$ and $4$). I tried to use summation signs but i does't seem to get simpler (is summation even usable in proportional logic?). I know the basics of induction by heart, but I don't know how to extend them to so many cases. Can someone show me how?

Ernie060
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Kripke Platek
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1 Answers1

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Since every formula is built from those operations, you should show that those operations preserve the equivalence. This is done by induction on the degree of a formula.

If $A_1$ is of degree $k$, then your base case is a formula containing $A_1$ as a part and of degree $k$. The only possible formula under those conditions is $A_1$ itself, so $A_1\equiv A_2$ is the base case, as you said.

Now suppose we have the result for all formulas containing $A_1$ as a part and of degree up to $n$ with $n\ge k$. Consider a formula $C$ containing $A_1$ as a part and of degree $n+1$. Since $C$ is of degree $n+1>n\ge k$ there is at least one propositional connective that doesn't belong to $A_1$.

Here is where you have to consider the cases where that extra propositional connective could be $\lnot$, $\lor$, $\land$, $\Rightarrow$, $\Leftarrow$, $\Leftrightarrow$; and separate $C$ is two different subformulas (one of them containing $A_1$ as a part, since we didn't break $A_1$). You apply your induction hypothesis in that subformula (we can because it is of degree $l$ with $k\le l\le n$, with the first inequality holding because it contains $A_1$) and replace $A_1$ for $A_2$ in that subformula.

Finally, justify that the whole formula, $C$, is equivalent to the one obtained replacing $A_1$ for $A_2$ knowing the equivalence of the subformula and using the propositional connective you're considering. Once you've considered all cases, you're done.

Are you able to do it like this?

Darsen
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  • Could you do it with that information? – Darsen Oct 08 '20 at 22:44
  • Well, It seems like a formula with degree n can have different ways of arranging parenthesis .Like : (A1 and B) or (A1 and C) seems to be non-reachable with the induction method.Is there something I am missing (This is bugging me a lot) – Kripke Platek Oct 20 '20 at 15:55
  • Also I didn't quite understand the part "Now suppose we have the result for all formulas containing A1 as a part and of degree up to n with n≥k." – Kripke Platek Oct 20 '20 at 16:11
  • You want to prove that the result is true for any formula $C$ containing $A_1$ as a part. You have proved the base case, which is degree $k$, so now you want to prove: if every formula up to degree $n$ (for some $n\ge k$) that contains $A_1$ as a part satisfies the statement, then a formula of degree $n+1$ that contains $A_1$ as a part satisfies it too. This is the induction step in strong induction. – Darsen Oct 20 '20 at 18:30
  • And yes: a formula of degree $n+1$ that contains $A_1$ as a part could be arranged in many ways; but you can consider each case. Since $A_1$ is a part of that formula by hypothesis, $A_1$ has to be connected to the rest of the formula by some propositional connective. It could be $\lnot A_1$, $B\lor A_1$, $B\Rightarrow A_1$... But it has to be some of them. Now, if you "remove" the propositional connective you're left with a formula of degree $\le n$, so you can apply you induction hypothesis (this is why we use strong induction), and then consider the equivalence adding back the connective. – Darsen Oct 20 '20 at 18:35
  • But for me , This method only seems to work for formulas like: (((A1 V B) ∧ C ) ∧ (¬A1) ) And not for formulas like this : ((A1 V B) ∧ (C ∧ (¬A1))) .(Which one is even the base A1? ).I Guess now I have to consider every parenthesis case and treat it like a induction + combinatorial problem ? (Maybe I am over complicating things here ) – Kripke Platek Oct 20 '20 at 19:03
  • You are overcomplicating things indeed. You're right about the type of formula though, I should have done an extra indication. But the strong induction remains: if $C$ is a formula of degree $n+1$ that contains $A_1$, then $A_1$ or $\lnot A_1$ has to be part of a subformula or several subformulas of $C$. If the only subformula is $C$ itself, then $C$ is of the first type you wrote, and you can do it with the propositional connectives as we talked. – Darsen Oct 20 '20 at 19:33
  • If it's not $C$ then $A_1$ is part of subformulas of degree $\le n$, so you can apply the induction hypothesis to each subformula. To end, since each subformula is equivalent to the one resulting by exchaging $A_2$ for $A_1$, their truth values don't change in any case, so they work the same as subformulas of $C$, and thus $C$ is equivalent to the formula resulting by exchaging $A_2$ for $A_1$. – Darsen Oct 20 '20 at 19:33