Is it always needed to go through bi-conditionals in separated steps when proving equivalence? Proving $A \subseteq B \Leftrightarrow B' \subseteq A'$

Question

Prove that if $A$ and $B$ are subsets of U, then $A \subseteq B \Leftrightarrow U-B \subseteq U-A$

$$U - B \subseteq U-A \Leftrightarrow (\forall x)(x \in U-B \Rightarrow x \in U - A)$$ $$U - B \subseteq U-A \Leftrightarrow (\forall x)((x \in U \land x \notin B) \Rightarrow (x \in U \land x \notin A))$$ $$U - B \subseteq U-A \Leftrightarrow (\forall x)((x \notin U \lor x \in B) \lor (x \in U \land x \notin A))$$ $$U - B \subseteq U-A \Leftrightarrow (\forall x)((x \in U \lor x \notin U \lor x \in B) \land (x \notin A \lor x \notin U \lor x \in B))$$ $$U - B \subseteq U-A \Leftrightarrow (\forall x)( \top \land (x \notin U \lor (x \notin A \lor x \in B)))$$ $$U - B \subseteq U-A \Leftrightarrow (\forall x)(x \notin U \lor (x \in A \Rightarrow x \in B))$$ $$U - B \subseteq U-A \Leftrightarrow (\forall x)(x \in U \Rightarrow (x \in A \Rightarrow x \in B))$$ $$U - B \subseteq U-A \Leftrightarrow (\forall x)((x \in U \land x \in A) \Rightarrow x \in B)$$ And as we have $A$ being a subset of $U$ we have also the following implication: $$(\forall x)(x \in A \Rightarrow x\in U) \Rightarrow (\forall x)(x \in A \Leftrightarrow x \in A \land x \in U)$$ Then we end with: $$U - B \subseteq U-A \Leftrightarrow (\forall x)(x \in A \Rightarrow x \in B)$$ $$U-B \subseteq U - A \Leftrightarrow A \subseteq B$$

My doubts are if I can consider tautology to do simplifications and if it is right to prove equivalence without going through both bi-conditionals in separate steps, and if the solution is correct.

Answer 1

Here is a simple direct proof.

A subset B iff
for all x, (x in A implies x in B) iff
for all x, (x not in B implies x not in A) iff
B' subset A'

What you have written is excessively complex and does not use well formed formulas (mathematical incoherent).

Is a temporary universal set some sort of neomath oxymoron?