Let $A$, $B$, $C$ be statements. Given $A\iff(B∨C)$.
I am trying to prove (if possible) that the implication $A \Rightarrow \neg B$ is false.
The motivation for me to do this is that I am thinking of eliminating the ‘‘redundant’’ cases in an implication (by that I mean, e.g. the implication $\forall x\in \mathbb{R},x^2=1\Rightarrow(x=1∨x=2∨x=-1)$ is true but $x=2$ is redundant). The solution I come up with is to prove for both directions of the implication as above so that the equivalence should force both $B$ and $C$ not to be redundant. To verify this, I would like to prove that with this equivalence, if it is the case of $A$, then it is not the case that, say, $B$ could never happen. And, for me, this is the same as to prove that given $A\iff(B∨C)$, $A \Rightarrow \neg B$ is false.
The argument I conduct is as follows:
Assume $B \Rightarrow \neg A$ is true. Since we have $B \Rightarrow (B∨C)\Rightarrow A$, by contradiction $B \Rightarrow \neg A$ is false. And since $A \Rightarrow \neg B$ is the contrapositive of $B \Rightarrow \neg A$, we have $A \Rightarrow \neg B$ is false.
However, maybe it is due to a lack of quantifiers or maybe because of the way I use that contraposition/contradiction, or some other things: I find my proof really confusing in term of its logic flow, and I have a feeling that somethings are missing/misused so that there are some flaws in it that are needed to be fixed/completed.
Could anyone please clarity for me?
Well, if $A\Leftrightarrow B\vee C$, then $A\Rightarrow \neg B$ becomes $B\vee C\Rightarrow \neg B$, which is not a tautology by taking $B,C$ to be true. Then the premise is true, but the conclusion is false.