4.2 THE COMPLETENESS THEOREM: (2) If A theory $\mathbf{T}$ has a model, then it is consistent.

4.2 THE COMPLETENESS THEOREM

(2) If A theory $\mathbf{T}$ has a model, then it is
consistent.

Proof

Suppose that $\mathbf{T}$ has a mode $\mathbf{\alpha}$.

By definition of valid(P19), a formula $\mathbf{A}$ of $\mathbf{L}$ is valid
in $\mathbf{\alpha}$ if $\mathbf{\alpha(A‘)=T}$ for every closed formula
$\mathbf{A‘}$ of $\mathbf{A}$,we get a closed formula $\mathbf{A \& \neg A}$
of $\mathbf{L}$ is not valid in $\mathbf{\alpha}$ if $\mathbf{\alpha(A \&
\neg A)=F}$.

By validity theorem(P23), every theorem of theory T is valid in T, we get if
$\mathbf{A \& \neg A}$ is not valid in $\mathbf{\alpha}$, then the formula
$\mathbf{A \& \neg A}$ is not theorem.

By definition of consistent(P42), a theory $\mathbf{T}$ is consistent if not
every formula of $\mathbf{T}$ is a theorem of $\mathbf{T}$, we get if the
formula $\mathbf{A \& \neg A}$ is not theorem, then a theory $\mathbf{T}$ is
consistent.

i.e.,

$\mathbf{\alpha(A \& \neg A)=F} \Rightarrow
\mathbf{A \& \neg A}$ is not valid in $\mathbf{\alpha}$.

$\mathbf{A \& \neg A}$ is not valid in
$\mathbf{\alpha} \Rightarrow \mathbf{A \& \neg A}$ is not theorem.

$\mathbf{A \& \neg A}$ is not theorem
$\Rightarrow T$ is consistent.

4.2 THE COMPLETENESS THEOREM: (2) If A theory $\mathbf{T}$ has a
model, then it is consistent.

时间: 2024-10-19 21:12:45

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