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Simplify $(A\implies B)\implies(A\implies C)$ as much as possible.

I've converted the implications so that $$(\neg A\vee B)\implies(\neg A\vee C)$$ and then $$\neg(\neg A\vee B)\vee(\neg A\vee C)$$ but after that I don't know what more I can do. I've tried applying De Morgan in a hope to reduce it to only one $A$, but I keep going in circles.

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If you go on with De Morgan laws you get $$ (A\land \lnot B)\lor(\lnot A\lor C) $$ that can be reordered as $$ \bigl((A\land \lnot B)\lor\lnot A\bigr)\lor C $$ Apply distributivity to the expression $(A\land \lnot B)\lor\lnot A$, that gives $(A\lor\lnot A)\land(\lnot B\lor\lnot A)$; the first term can be removed, being true, so we remain with $$ (\lnot B\lor\lnot A)\lor C $$ and, again with De Morgan, we get $(A\land B)\Rightarrow C$

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