When dealing with the question -> 1. Show that ( → ) → ( → ) is logically equivalent to → using logical equivalence rules. Name each rule. You will get no credit for any other type of solution, such as a solution by truth table.
I understand I need to go through and use the laws, but I'm not entirely sure what I'm supposed to be doing. Am I supposed to go through and alter the p & q to fit into the other laws? (Such as De Morgan, Communtative, ect)
Such as, ≡ ¬ (¬p ∨ q) ∨ (¬q∨p) [DeMorgan law? P.s I don't know if I did this one right]
As stated your question is: Show that ( → ) → ( → ) is logically equivalent to → using logical equivalence rules. Name each rule. You will get no credit for any other type of solution, such as a solution by truth table.
Here you two show equivalence of two logical statements using the rules, naming each of the rule. Also use of truth table to prove equivalence is prohibited.
Proof of equivalence: $( → ) → ( → ) \equiv → $
LHS:
$(p → q) → (q → p) = (\sim p \lor q) → (\sim q \lor p)$. $\quad[Def. of implication]$
$ \Rightarrow(\sim(\sim p \lor q) \lor (\sim q \lor p)\,)$ . $\quad[Def. of implication]$
$ \Rightarrow (p\,\land \sim q) \lor (\sim q \lor p)\,)$ . $\quad[DeMorgan's laws]$
$ \Rightarrow (∼q∨p)$ . $\quad[Associative\,\&\,Distributive laws]$
$ \Rightarrow (∼q∨p) = (q → p)$ = RHS. $\quad[Def. of implication]$
$\therefore LHS = RHS$.
Hence Proved.