If $\beta$ is an $\mathcal{L}$ formula and $\beta_P$ is a tautology then $\beta $ is valid I am reading A Friendly Introduction to Mathematical Logic by Christopher Leary and Lars Kristiansen. Let $\beta$ be a $\mathcal{L}$-formula in first order language $\mathcal{L}$. In the first line of Page 52, the authors claim that if $\beta_P$, a formula in propositional logic, is a tautology then $\beta$ is valid.
$\beta_P$ is obtained from $\beta$ in the following fashion (which is written is Page 51):

*

*Find all subformulas of $\beta$ of the form $\forall x\alpha$ that are not in the scope of another quantifier. Replace them with propositional variables in a systematic fashion. This means that if $\forall yQ(y,c)$ appears twice in $\beta$, it is replaced by same letter both times, and distinct subformulas are replaced with distinct letters.


*Find all atomic formulas that remain, and replace them systematically with new propositional variables.


*At this point, $\beta$ will have been replaced with a propositional formula $\beta_{P}$.
I was trying to prove this by induction on the structure of formula $\beta$ but I got stuck in the base case. Here's my attempt: Let $\mathfrak{U}$ be any $\mathcal{L}$-structure and $s$ be any variable assignment function to $\mathfrak{U}$. Suppose that $\beta\equiv=t_1t_2$ (written in Polish notation) where $t_1 , t_2$ are terms of the language $\mathcal{L}$. Then $\beta_P$ is $A$ where $A$ is a propositional variable. Let's assume that $\beta _P$ is a tautology. We need to prove that $\beta$ is valid, i.e., $\mathfrak{U}\models \beta [s]$.
Here's my question: Since $\beta_P$ is a tautology, are not allowed to make the substitution $A=F$ because if we do, then $\beta_P$ would not longer be a tautology. How do we remedy this situation?
 A: We need a preliminary Lemma in order to define a "procedure" that assigns to every interpretation $\mathfrak A$ and variable assignment function $s$ a boolean valuation $v$ such that:

(1) $\mathfrak A,s \vDash \beta \text {  iff  } v(\beta_{\text P})= \text T$.

The idea is to consider the rule for producing $\beta_{\text P}$ from $\beta$ above and let $P_i$ the propositional variable corresponding to subformula $Q_i$, i.e. an atom or subformula $\forall x \alpha$.
We define:

$v(P_i)= \text T \text { when } \mathfrak A,s \vDash Q_i, \text { and } v(P_i)= \text F$ otherwise.

And then we use the clauses for the satisfaction relation [Def.1.7.4, page 29] to prove (1) by induction, considering that we have three "atomic" cases: (1) $=t_1t_2$, (2) $Rt_1 \ldots t_n$ and (3) $∀xα$, and only two compound cases: (4) $¬ \varphi$ and $\varphi ∨ \psi$.
Having proved the Lemma, we can use it to prove the main result.
Let $\beta$ a formula and let $\mathfrak A$ and $s$ whatever.
We compute the corresponding $v$. But $\beta_{\text P}$ is a tautology; thus: $v(\beta_{\text P})=\text T$, that implies: $\mathfrak A,s \vDash \beta$.
A: As I see it, the problem is that the final $\beta_P$ might be a tautology while its parts you obtain in your induction might not be.
You are right that the partial $\beta_P$ you get from e.g. $t_1 = t_2$ would be $A$ and thus not a tautology, but the complete $\beta_P$ might be $A \lor \neg A$, which is a tautology.
