1. ## Fol 4

This is another exercise problem (Exercise 3 of Section 2.4) of Enderton's "A mathematical introduction to logic".

(a) Let $A$ be a structure and let $s:V \rightarrow |A|$. Define a truth assignment $v$ on the set of prime formulas by $v(\alpha)=T$ iff $\models_A \alpha[s]$. Show that for any formula (prime or not), $\bar{v}(\alpha) = T$ iff $\models_A \alpha[s]$.

(b) Conclude that if $\Gamma$ tautologically implies $\phi$, then $\Gamma$ logically implies $\phi$.

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What I have tried so far is

For (a), if $\alpha$ is a prime formula, nothing needs to be done. If $\alpha$ is not a prime formula, two cases have to be considered, i.e., $\alpha = \neg \beta$ or $\alpha = (\beta \rightarrow \gamma)$, where $\beta$ and $\gamma$ are both atomic formulas.

For (b), if $\bar{v}(\psi) = T$ for every $\psi \in \Gamma$, then $\bar{v}(\phi)=T$ by assumption. Then I need to show if $\models_A \psi[s]$, then $\models_A \phi[s]$. I am stuck here.

Any help will be appreciated.

2. ## Re: Fol 4

Originally Posted by logics
(a) Let $A$ be a structure and let $s:V \rightarrow |A|$. Define a truth assignment $v$ on the set of prime formulas by $v(\alpha)=T$ iff $\models_A \alpha[s]$. Show that for any formula (prime or not), $\bar{v}(\alpha) = T$ iff $\models_A \alpha[s]$.
It seems that only quantifier-free formulas are considered here.

Originally Posted by logics
For (a), if $\alpha$ is a prime formula, nothing needs to be done. If $\alpha$ is not a prime formula, two cases have to be considered, i.e., $\alpha = \neg \beta$ or $\alpha = (\beta \rightarrow \gamma)$, where $\beta$ and $\gamma$ are both atomic formulas.
You cannot assume that $\beta$ and $\gamma$ are atomic formulas (does it mean the same as prime formulas?) in the induction step. E.g., $(\alpha\to \alpha)\to \alpha$ where $\alpha$ is a prime formula has the form $\beta\to\gamma$, but $\beta$ is not prime. I assume you can finish the definition and the proof that $\bar{v}(\alpha) = T$ iff $\models_A \alpha[s]$.

Originally Posted by logics
For (b), if $\bar{v}(\psi) = T$ for every $\psi \in \Gamma$, then $\bar{v}(\phi)=T$ by assumption.
I believe this holds for any truth assignment v. Call this fact (*).

Originally Posted by logics
Then I need to show if $\models_A \psi[s]$, then $\models_A \phi[s]$.
Suppose $\models_A\psi[s]$ for every $\psi$ in $\Gamma$. Let $v_{A,s}$ be the truth assignment constructed from $A$ and $s$ as described in (a). Then $\bar{v}_{A,s}(\psi)=T$ by (a), so by putting $v=v_{A,s}$ in (*) we have $\bar{v}_{A,s}(\phi)=T$, which again by (a) implies $\models_A\phi[s]$.