1- ((¬P ∧ ¬Q) ∨ (P ∨ R) ∨ (Q ∨ R))∨ R
2- (¬P ∧ ¬Q) ∨ (P ∨ R) ∨ (Q ∨ R)∨ R
3- (¬P ∧ ¬Q) ∨ (P ∨ R) ∨ (Q ∨ R) ∨ (R ∨ R)
Q:Why did we remove the double parenthesis after step number 1?
Also, my thinking process was that I could distribute the R outside the double parenthesis over the sub-groups inside the double parenthesis (Distribute it over the whole thing) and not only distribute the R outside the double parenthesis over (Q ∨ R) only.
*Note: I am new to Discrete-Mathematics and to this site. I would very appreciate some tips on what to improve and learn, but be gentle with me since I am like still a newbie.
Something isn't right:
1 $\neg ((\neg P \land \neg Q) \lor (P \lor R) \lor (Q \lor R)) \lor R$
is not equivalent to
2 $\neg (\neg P \land \neg Q) \lor (P \lor R) \lor (Q \lor R) \lor R$
Rather, it is equivalent to:
2 $(\neg (\neg P \land \neg Q) \land \neg (P \lor R) \land \neg (Q \lor R)) \lor R$
And notice that with that, you can not drop parentheses.