Show that if $\vdash A_{m+1}$, then $ A_1,...,A_m \vdash B$ if and only if $ A_1,...,A_m,A_{m+1} \vdash B$.

Question

This question concerns propositional logic. The $\vdash$ symbol refers to deductibility. How do I prove this without using the fact that propositional logic is consistent? Thanks.

Show that if $\vdash A_{m+1}$, then $ A_1,...,A_m \vdash B$ if and only if $ A_1,...,A_m,A_{m+1} \vdash B$.

Answer 1

The derivability relation: $\vdash$ has the following property (also called Weakening):

if $\Gamma \vdash \varphi$ and $\Gamma \subseteq \Gamma'$, then $\Gamma' \vdash \varphi$.

We can use it for half of the needed result:

if $A_1,\ldots,A_m⊢B$, we can add an additional premise to get: $A_1,\ldots,A_m,A_{m+1} ⊢ B$.

For the other part, we can use the Deduction Theorem:

from $A_1,\ldots,A_m,A_{m+1}⊢B$, we can derive: $A_1,\ldots,A_m ⊢ A_{m+1} → B$.

Then, we can use $⊢A_{m+1}$ and modus ponens to get: $A_1,\ldots,A_m⊢B$.