i am trying to figure whether the following claims are equivalent:
1)$\Sigma$ is consistent and complete
2)for 2 sets $\gamma_1$ $\gamma_2$: $\forall ( y_1\in \gamma_1, y_2\in \gamma_2)$ $\Sigma \vDash (y_1 \lor \gamma_2) \iff \forall y_1 \in \gamma_1$ if $\Sigma \not \vdash y_1$ then for every $y_2 \in \gamma_2$ occurs $\Sigma \vdash y_2$
my attempt:
$\Sigma$ is considered to be consistent iff it does not contain any contradiction, and it is complete if we can express each of its members inside a truth table as combination of the other using boolean expressions. however, because of $\Sigma \not \vdash y_1$ then for every $y_2 \in \gamma_2$ occurs $\Sigma \vdash y_2$, if we exchange the $\lor$ to and $\rightarrow$ we get $\Sigma \vDash \lnot y_1 \rightarrow y_2$.
so i think that the claims are equivalent and correct.
please correct me if i've done something wrong. thank you very much