Prove $A cup (cap B)=cap{Acup b: b in B}$

x in A $cup$ $cap$B iff
x in A or for all S in B, x in B iff
for all S in B, (x in A or x in S) iff
for all S in B, x in A $cup$ S iff
x in $cap${ A $cup$ S : S in B }

What you write as $cap{Acup b:bin B}$, I suspect you mean ${x:forall bin B~.xin Acup b}$, or simply $bigcap_{bin B} (Acup b)$

(Similarly $bigcap B$ appears to be a shorthand for ${bigcap}_{bin B} b$, which is ${x:forall bin B~.xin b}$ )

Thus you seek to prove: $$begin{align}Acup bigcap B &= {x: xin Alor xinbigcap B} \[1ex] &={x:xin Alor (forall bin B~.xin b)}\[1ex]&={x:forall bin B~.(xin Alor xin b)}\[1ex]&={x:forall bin B~.xin (Acup b)}\[1ex]&={bigcap}_{bin B}(Acup b)end{align}$$

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Let $x in cap{A cup b: b in B}$, then $forall a, a in Acup b , xin a$ then $forall a, a in A $ $vee$ $a in b $ then $forall a, a in A, x in a $ $vee$ $forall a, a in b, x in a$ then $x in A$ $vee$ $forall a in B, x in a $ this is beacuase $a in b wedge b in B Rightarrow a in B$ and $x in a wedge a in A Rightarrow x in A$; then $x in A cup (cap B)$

No, $xin awedge ain A$ does not imply $xin A$.   Moreover you are confusing your $a$, $b$, and $x$ elements.   We just need $x$ and $b$ — the arbitrary $x$ is assumed to be in the thing, and we discuss all $b$ that are in $B$.

• Take an arbitrary $x$ with the assumption that $xin bigcap_{bin B}(Acap b)$. That is to say: $forall bin B~.(xin Acup b)$. Therefore either $xin A$ or if otherwise you can show $forall bin B~.(xin b)$.   Hence $xin Acupbigcap B$.

Thereby proving : $bigcap_{bin B}(Acup b)subseteq Acupbigcap B$.

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The converse may be proven through a proof by cases.

• Take an arbitrary $x$ with the assumption that $xin Acupbigcap B$. That is to say: $xin A$ or $xin bigcap B$.
  • In the case of $xin A$, there you can show: $forall bin B~.xin (Acup b)$.
  • In the case of $xin bigcap B$ [that is to say $forall bin B~.xin b$], there you can show: $forall bin B~.xin (Acup b)$.
• Therefore $forall bin B~.xin (Acup b)$ which is to say: $xinbigcap_{bin B}(Acup b)$

Thereby proving : $bigcap_{bin B}(Acup b)supseteq Acupbigcap B$.

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Together proving: $bigcap_{bin B}(Acup b) = Acupbigcap B$.

$blacksquare$