Let’s say I have some formal theory $T$, and I have a statement $A$. Let $L$ be an extension of $T$ s.t. $$ L = T \cup \{A\} $$ If we show that $T$ is consistent, can we show that $L$ is consistent so long as $$ T \not\vdash \lnot A $$ That is, say if we assume that $T$ is consistent, and we have some model of $T$ where $A$ is true, does this necessarily imply the consistency of $L$ given that $T$ is consistent by assumption?
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Yes; we have that if $Γ \cup { \varphi }$ is inconsistent, then $Γ \vdash \lnot \varphi$. And thus, if $Γ \nvdash \lnot \varphi$, then $Γ \cup { \varphi }$ is consistent. – Mauro ALLEGRANZA Nov 16 '21 at 08:58
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And from a semantical point of view, consistent means satisfiable; thus, if we have a model $M$ satisfying $T$ and we have that $A$ is true in $M$, then this shows that $T \cup { A }$ is satisfiable. – Mauro ALLEGRANZA Nov 16 '21 at 09:00
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See also this post – Mauro ALLEGRANZA Nov 16 '21 at 09:09
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Yes, this follows from the deduction theorem. The following version of the theorem is excerpted from Shoenfield's Mathematical Logic (with notations changed to fit the question):
Let $A$ be a statement. For every formula $B$ of $T$, $T \vdash A \to B$ iff $B$ is a theorem of $T \cup \{A\}$.
If $L$ is inconsistent, then $\perp$ is a theorem of $T \cup \{A\}$, so $T \vdash A \to \perp$, iff $T \vdash \neg A \;\vee \perp$, iff $T \vdash \neg A$.
Clement Yung
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