If the union of two theories is not consistent there exists a sentence $P$ for which one of them implies $P$ and the other one implies $\lnot P$.
Can someone please check if this is true or not?
I used to prove this by contradiction. I assumed for every sentence $P$, if $T_1$ implies $P$, then $T_2$ satisfies $P$ or $T_2$ doesn't imply $P$ but never $T_2$ implies $\lnot P$. So i think there exists a model of $T_2$ as a substructure of a model of $T_1$( i'm not sure if this is true... I just say it intuitively ), so I take that model of $T_1$ as a model of $T_1 \cup T_2$. Since the union doesn't have a model so there exists a sentence $P$ such that $T_1$ implies $P$ and $T_2$ implies the $\lnot P$.