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

本文阐述了数学逻辑中完备性定理的一个关键方面：如果一个理论拥有至少一个模型，则该理论是一致的。通过定义有效性和一致性的概念，证明了在给定模型下矛盾公式既非有效也非定理，从而得出理论的一致性。

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.

转载于:https://www.cnblogs.com/mathematicallogic/p/3763695.html