Proving the distribution of union operation on the intersection of two sets. I have to prove $$ \left(A \cap B \right) \cup C = \left(A \cup C \right) \cap \left(B \cup C\right)$$
My attempt is the usual one : 
$$ \textrm{Let}~ x \in \left[ 
\left (
A \cap B 
\right)
\cup C
\right] $$
$$\implies x \in \left(A \cap B \right) ~\textrm{or}~ x \in C $$
$$\implies \left( x \in A ~\textrm{and}~ x \in B \right) ~\textrm{or}~ x \in C $$
The last line can be written in words as " x belongs to A and simultaneously to B or x belongs to C". Now, how to proceed logically? I'm in need of a detailed answer explaining how interpret and manipulate the statement " x belongs to A and simultaneously to B or x belongs to C ". 
Thank you!  
 A: Just reason with words instead of formulae (otherwise you reduce it to a similar statement in propositional logic):
Let $x \in (A \cap B) \cup C$. This means we know that $x \in C$ or $x \in A \cap B$. If $x \in C$ then $x \in A \cup C$ and $x \in B \cup C$ too. So
$x \in (A \cup C) \cap (B \cup C)$, in that case. Otherwise $x \in A \cap B$ and so $x \in A$ (so also in $A \cup C$) and $x \in B$ (so also in $B \cup C$, and again $x \in (A \cup C) \cap (B \cup C)$.
The other inclusion: let $x \in (A \cup C) \cap (B \cup C)$. So $x \in A \cup C$ and also $x \in B \cup C$. If $x \in C$ we are done as then $x \in (A \cap B) \cup C$ a fortiori. So assume $x \notin C$ instead. Then $x \in A$ (otherwise $x \notin A \cup C$) and $x \in B$ (or else $x \notin B \cup C$), so $x \in A \cap B$ and hence $x \in (A \cap B) \cup C$. These two cases prove the other inclusion.
The two inclusions together show equality of sets.
A: $$(A∩B)∪C≡\{x:x∈(A∩B)∪C\}$$$$≡\{x:(x∈A∩B) ∨ x∈C\}$$$$≡\{x:(x∈A ∧ x∈B) ∨ x∈C\}$$$$≡\{x:(x∈A ∨ x∈C) ∧ (x∈B∨x∈C) \}$$$$≡\{x:(x∈A∪C)∩(x∈B∪C)\}$$$$≡\{x:x∈(A∪C)∩(B∪C)\}$$
A: Prove distributivity of 'or' over 'and'.
(P and Q) or R .iff. (P or R) and (Q or R)  
Hints.
Left to right.
(P and Q) or R .implies. P or R   
Right to left.
Two cases:  R and not-R.  
A: Equality amounts to reciprocal inclusion. 
Below is shown the inclusion of the LHS set in the RHS set. 

