Below is the proof for the following proposition:

(p ⇒q) ⇔ (¬q⇒¬p)

For each step in the proof indicate which law is used.

We start with:

p⇒q

Step 1: ¬p∨q

Step 2:q∨¬p

Step 3: ¬(¬q)∨¬p

Step 4: ¬q⇒ ¬p

What law belongs to what step?

Law 1:(a∨ b) ⇔ (b∨ a)(commutativity of∨)

Law 2:¬(¬a)⇔a(law of negation)

Law 3:(a⇒ b)⇔ (¬a∨ b)(law of implication)

Law 4:(a⇔b)⇔ ((a⇒b)∧ (b⇒a))(law of equivalence)I really have no clue where to start here ... so maybe someone can give me a hand ...