Definitions: Let $\Lambda$ be a set of logical axioms and $\Gamma$ be a sets of well-formed formulas (in propositional logic). We say that $\Gamma\cup\Lambda$ tautologically implies $\varphi$ if for every truth assignment satisfying every member of $\Gamma\cup\Lambda$ also satisfies $\varphi.$
We say that $\Gamma\vdash \varphi$ if there exists a finite sequence $\langle a_0,\dots,a_n\rangle$ in $\Gamma$ such that $a_n=\varphi$ and for each $k\leq n,$ either
$(1)$ $a_k$ is in $\Gamma\cup\Lambda,$ or
$(2)$ $a_k$ is obtained by modus ponens from two earlier formulas in the sequence, that is, for some $i$ and $j$ less than $k,$ $a_j$ is $a_i\to a_k.$
I'm reading Enderton's logic book. In page $115,$ he stated the following theorem.
Theorem $24$B: $\Gamma\vdash \varphi$ iff $\Gamma\cup \Lambda$ tautologically implies $\varphi$.
The author provided the following proof to $(\Rightarrow)$ direction:
This depends on the obvious fact that $\{\alpha,\alpha\to\beta\}$ tautologically implies $\beta.$ Suppose that we have a truth assignment $v$ satisfies every member of $\Gamma\cup \Lambda.$ By induction we can see that $v$ satisfies any theorem of $\Gamma.$ The inductive step uses exactly the above-mentioned fact obvious fact.
Question: I fail to prove the bolded sentence. In particular, I do not see how to use induction.
My attempt: Let $v$ be a truth assignment that satisfies all members of $\Gamma\cup\Lambda.$ By assumption, let $\langle a_0,...,a_n\rangle$ be a finite sequence in $\Gamma$ such that $a_n=\varphi.$
For each $k\leq n,$ if $a_k$ is in $\Gamma\cup\Lambda,$ then $v$ will satisfy $a_k.$
We use induction to show that if $a_k$ is obtained by modus ponens from two earlier formulas, say $a_i$ and $a_j$ for $i$ and $j$ less than $k,$ then $v$ will satisfy $a_k.$
Let $a_k$ be the first formula that is obtained by modus ponens from $a_i$ and $a_j$ where $i$ and $j$ are less than $k.$ Then $a_i$ and $a_j$ are elements of $\Gamma\cup\Lambda.$ Therefore, $v$ will satisfy $a_i$ and $a_j.$ By modus ponen, $v$ will satisfy $a_k.$
Our inductive hypothesis is that if $a_k$ is obtained by modus ponens from two earlier formulas, then $v$ will satisfy $a_k.$
Suppose that $a_{k+1}$ is the next formula obtained by modus ponens from two earlier formulas, say $a_i$ and $a_j$, where $i$ and $j$ are less than $k+1.$ By inductive hypothesis, $v$ will satisfy both $a_i$ and $a_j.$ Therefore, $v$ will satisfy $a_{k+1}.$
Is my proof correct?