Prove: (M − N) ∪ (N − P) ⊆ (M ∪ N) − (M ∩ N ∩ P) are there mistakes or is my proof incomplete?

Question

$$(M\setminus N) ∪ (N\setminus P)=$$

$$(M\cap N^c)\cup(N\cap P^c)=$$

$$(M\cup N)\cap (M\cup P^c)\cap (N^c\cup P^c)\cap (N\cup N^c)=$$

$$(M\cup N)\setminus[(M \cup P^c)\cap (N^c\cup P^c)]^c=$$

$$(M\cup N)\setminus[(M\cup P^c)^c\cup(N^c\cup P^c)^c]=$$

$$(M ∪ N)\setminus[(M^c\cap P)\cup(N\cap P)]=$$

$$(M ∪ N)\setminus[P\cap(M^c\cup N)]=$$

$$(M ∪ N)\setminus[P\cap(M^c\cup N)\cap (M\cup M^c) ]=$$

$$(M\cup N)\setminus[P\cap(M^c\cup(M\cap N)]=$$

$$(M ∪ N)\setminus[(P\cap M^c)\cup (P\cap M\cap N)]=$$

I am worried about mistakes and logical leaps. Is this proof incomplete? At the end, do I need to show/how do I show that $${(M\cup N)\setminus[(P\cap M^c)\cup (P \cap M\cap N) ]}\subseteq(M\cup N)\setminus(M\cap N\cap P) $$ Are there any other mistakes. Any help/verification/general insight is greatly appreciated.

Answer 1

You can use contraposition now: $$(M\cup N)\setminus[(P\cap M^c)\cup (P \cap M\cap N) ]\equiv(M\cup N)\cap[(P\cap M^c)\cup (P \cap M\cap N) ]^c$$ $$A\subseteq B\equiv B^c\subseteq A^c$$

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explanation: $$(P \cap M\cap N)\subseteq[(P\cap M^c)\cup (P \cap M\cap N)]\equiv[(P\cap M^c)\cup (P \cap M\cap N)]^c\subseteq(P\cap M\cap N)^c$$

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extra:

$$A\setminus(B\cup C)\equiv A\cap(B\cup C)^c\equiv\underbrace{A\cap B^c\cap C^c}_{X}\equiv X$$ $$A\setminus C\equiv A\cap C^c\equiv A\cap C^c\cap(B^c\cup B)\equiv\underbrace{(A\cap B^c\cap C^c)}_{X}\cup\underbrace{(A\cap B\cap C^c)}_{Y}\equiv X\cup Y$$ $$\implies X\subseteq X\cup Y$$