How can I prove the generalized De Morgan laws using mathematical writing, and not induction (I am not allowed).
This is what I concluded:
$\bigcap_{i\in I}A_i' = (\bigcup_{i\in I}(A_i)).$ and its opposite:
$\bigcup_{i\in I}A_i' = (\bigcap_{i\in I}(A_i)).$
I know how to prove it using induction, but I am not allowed unfortunately. I am requested to prove only using logic (mathematical writing).
This is how I've generalized it: $\bigcup_{i\in I}A_i iff \exists \:i\left(i\in \:I\:\:∧\:x\in \:A_{_{_i}}\right)$
and the other one: $(\bigcap_{i\in I}(A_i)) iff \forall i(i\in I\:->\:X\in A_i)$
but I don't know how to prove it using logic for the generalized form.
Please show me how to do it correctly, I've written what I've already done.
$\endgroup$1 Answer
$\begingroup$We have that
\begin{align}x\in \bigcap_{i\in I}A_i'&\iff x\in A_i',\forall i\in I \\ &\iff x\notin A_i,\forall i\in I\\&\iff x\notin \bigcup_{i\in I} A_i\\&\iff x\in \left(\bigcup_{i\in I} A_i\right)' \end{align}
which shows $$\bigcap_{i\in I}A_i=\left(\bigcup_{i\in I} A_i\right)'.$$ Proceed in the same way to get the other equality.
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