4.2 THE COMPLETENESS THEOREM: (3) Lemma 1.

3. Lemma 1. if $\mathbf{T'}$ is an extension of $\mathbf{T}$, and $\mathbf{\alpha'}$ is a model of $\mathbf{T'}$, then the restriction of $\mathbf{\alpha'}$ to $\mathbf{L(T)}$ is a model of $\mathbf{T}$.

Proof.

Every closed formula $\mathbf{A}$ of $\mathbf{L(\alpha)}$ are valid in $\mathbf{\alpha'}$ are as in $\mathbf{\alpha}$, i.e., $\mathbf{\alpha(A)=\alpha'(A)}$

By the definition of closure, the closure of formula is closed, i.e.,$\{closure \ formula\}\subset\{closed \ formula\}$, we get

Closure of formula $\mathbf{A_1}$ of $\mathbf{L(\alpha)}$ are valid in $\mathbf{\alpha'}$ are as in $\mathbf{\alpha}$, i.e., $\mathbf{\alpha(A_1)=\alpha'(A_1)}$

By the corollary of the closure theorem, we get

Formula $\mathbf{A_2}$ of $\mathbf{L(\alpha)}$ are valid in $\mathbf{\alpha'}$ are as in $\mathbf{\alpha}$,

thus,

Nonlogical axioms of $\mathbf{L(\alpha)}$ are valid in $\mathbf{\alpha'}$ are as in $\mathbf{\alpha}$

By definition of extension, every nonlogical axiom of $\mathbf{T}$ is a theorem of $\mathbf{T'}$, and by validity theorem, a theorem of $\mathbf{T'}$ is valid in model of $\mathbf{T'}$, i.e., every nonlogical axiom of $\mathbf{T}$ is valid in model of $\mathbf{T'}$. By the condition of Lemma 1, $\mathbf{\alpha'}$ is a model of $\mathbf{T'}$, we get every nonlogical axiom of $\mathbf{T}$ is valid in $\mathbf{\alpha'}$ of $\mathbf{T'}$, thus ,

Every nonlogical axiom of $\mathbf{T}$ is valid in $\mathbf{\alpha}$, i.e., the restriction of $\mathbf{\alpha'}$ is a model of $\mathbf{T}$ by the definition of model.