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Suppose $Τ$ is an infinite set of propositional types and $\varphi$ a propositional type. Prove that if $Τ\vDash\varphi$, then a finite set $Τ_0\subseteq T$ exists, such that $Τ_0\vDash\varphi$.

I understand this is true based on my intuition but I cannot think of a strict mathematical way to justify it.

Any help would be appreciated.

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    $\begingroup$ See Compactness Theorem for Propositional Logic. $\endgroup$ – Mauro ALLEGRANZA Apr 18 at 15:29
  • $\begingroup$ @Chrysa Hi, are you familiar with truth trees? I can show you an easy and intuitive proof. $\endgroup$ – Simone Apr 19 at 20:42
  • $\begingroup$ I am familiar with those. I would appreciate it if you showed me this proof. $\endgroup$ – Chrysa Apr 21 at 14:11
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I was thinking about the proof using truth trees I promised you, but that would be a graphical nightmare of Tex, and I'm not sure I'm capable of providing it here. So I'm outlining a different proof instead; see if this convinces you, if not then I'll prove it using the method I promised you but I'm afraid it would have to be rather wordy and descriptive.

Since you're are familiar with truth trees then I presume that I don't have to persuade you about the fact that "$Z\vDash\varphi$" can be put as "$Z,\neg\varphi\vDash$" , (in words: "$Z,\neg\varphi$ is inconsistent"). So the claim you were trying to wrap your head around is a particular case of the Compactness Theorem for propositional logic which can be stated as:

if $Z\vDash$ then, for some finite $Z'$, included in $Z$, $Z\vDash$ .

A lemma $(L1)$: Let "$\Gamma\succcurlyeq$" mean: "for some finite $\Gamma'$, included in $\Gamma$, $\Gamma'\vDash$". For any sentence-letter $P_i$: If $\Gamma,P_i\succcurlyeq$ and $\Gamma,\neg P_i\succcurlyeq$ then $\Gamma\succcurlyeq$.

Proof: suppose $\Gamma,P_i\succcurlyeq$ and $\Gamma,\neg P_i\succcurlyeq$, then by definition for some finite $\Gamma'$ in $\Gamma$, we have $\Gamma',P_i\vDash$ and also for some finite $\Gamma''$ in $\Gamma$, we have $\Gamma'',\neg P_i\vDash$. But if $\Gamma'',\neg P_i\vDash$ then $\Gamma''\vDash P_i$, and together with $\Gamma',P_i\vDash$ then by Cutting we get $\Gamma',\Gamma''\vDash$. But $\Gamma',\Gamma''$ is also finite and included in $\Gamma$, thus the result follows $\Box$

Let $\Gamma$ name some formulae, possibly a coutable infinity of them, and let $P_1,P_2,P_3,...$ be a list, possibly infinite, of all the sentence-letters in $\Gamma$. Define recursively a series of formulae as such: $$\Gamma_0 = \Gamma,$$

$$\Gamma_{n+1} =\begin{cases}\Gamma_n, P_{n+1}\;\;\text{if}\;\;\Gamma_n,\neg P_{n+1}\,\succcurlyeq, \\ \Gamma_n,\neg P_{n+1}\;\;\text{otherwise}. \end{cases}$$

Assume $\Gamma\not\succcurlyeq$, to be read as "it's not the case that $\Gamma\succcurlyeq$", (in words: "$\Gamma$ finitely satisfiable", "all finite $\Gamma'$ in $\Gamma$ are consistent"). As an inductive hypothesis let $\Gamma_k\not\succcurlyeq$, hence, by $(L1)$, $\;$ $\Gamma_k,P_{k+1}\not\succcurlyeq$ $\;$ or $\;$ $\Gamma_k,\neg P_{k+1}\not\succcurlyeq$. But in both cases it means that $\Gamma_{k+1}\not\succcurlyeq$.

Now we define an interpretation $I$ by stipulating that, for each letter $P_n$ $$|P_n|_I=1\;\;\text{iff}\;\; P_n\;\text{is in}\; \Gamma_n.$$ Define $\pm P_i$ to be either $P_i$ or $\neg P_i$, and let $\phi$ be any formula in $\Gamma$, containing just the letters $P_i,...,P_j$. Then by our construction, for some choice of $\pm P_i$ we have $\pm P_i,...,\pm P_j$ $\,$ in $\,$ $\Gamma_j$, and $\pm P_i,...,\pm P_j$ is finite. So by the inductive result just above $\pm P_i,...,\pm P_j$ is also consistent. But then $\pm P_i,...,\pm P_j \vDash\phi$ (think about truth tables if you need convincing for it), and since, by construction of $\Gamma_j$, the premises are true then so is $\phi$; that is: $|\phi|_I=1$.

So all formulae $\phi_n$ (no matter how many) are true in the intepretation $I$, this means that $\Gamma$ cannot be inconsistent, or symbolically: $\Gamma\nvDash$. We just proved that if $\,\Gamma\not\succcurlyeq\,$ then $\,\Gamma\nvDash\,$, which is just the contrapositive of the theorem we're trying to prove: $$ \text{if} \;\; \Gamma\vDash \;\; \text{then} \;\; \Gamma\succcurlyeq.$$

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