De Morgan's laws
Part 1

Let x ∈ A ∩ B ¯ {\displaystyle x\in {\overline {A\cap B}}}. Then, x ∉ A ∩ B {\displaystyle x\not \in A\cap B}.

Because A ∩ B = { y | y ∈ A ∧ y ∈ B } {\displaystyle A\cap B=\{\,y\ |\ y\in A\wedge y\in B\,\}}, it must be the case that x ∉ A {\displaystyle x\not \in A} or x ∉ B {\displaystyle x\not \in B}.

If x ∉ A {\displaystyle x\not \in A}, then x ∈ A ¯ {\displaystyle x\in {\overline {A}}}, so x ∈ A ¯ ∪ B ¯ {\displaystyle x\in {\overline {A}}\cup {\overline {B}}}.

Similarly, if x ∉ B {\displaystyle x\not \in B}, then x ∈ B ¯ {\displaystyle x\in {\overline {B}}}, so x ∈ A ¯ ∪ B ¯ {\displaystyle x\in {\overline {A}}\cup {\overline {B}}}.

Thus, ∀ x (x ∈ A ∩ B ¯ ⟹ x ∈ A ¯ ∪ B ¯) {\displaystyle \forall x{\Big (}x\in {\overline {A\cap B}}\implies x\in {\overline {A}}\cup {\overline {B}}{\Big)}};

that is, A ∩ B ¯ ⊆ A ¯ ∪ B ¯ {\displaystyle {\overline {A\cap B}}\subseteq {\overline {A}}\cup {\overline {B}}}.
punctuation:no_comma, oulipo:avoid_letter[o], detectable_content:postscript