I need some help with the following demonstration, I am a bit stuck. The statement goes as follow:
There are two automatons $A=(\Sigma,s,F,s_0,F_A)$ and $B=(\Sigma,s,F,s_0,F_B)$ ($A$ and $B$ are identical, but only differ in their final sets). If we know that all the states of both are reachable from $s_0$.
How to proof that: If $L(A)\cap L(B)=\emptyset \Rightarrow F_A \cap F_B = \emptyset$
I am a bit bad at demonstrations, and I don't know where to start, I know that if the terminal states are different then the words accepted by both automatons would be different, therefore $F_A \cap F_B = \emptyset$ would be true.
Taking the contrapositive of your statement, it suffices to prove that $$ F_A \cap F_B \not= \emptyset \implies L(A)\cap L(B) \not= \emptyset $$ Let $q \in F_A \cap F_B$. Since $q$ is reachable from $s_0$, there is a path of label $u$ from $s_0$ to $q$. Since $q \in F_A \cap F_B$, this means that $u$ is accepted by $A$ and $B$, that is, $u \in L(A)\cap L(B)$. Thus $L(A)\cap L(B) \not= \emptyset$.