Correctness of sequent calculus

Question

I have a doubt here, does the left of <-> lead to 4 paths? Could you please verify this?

Answer 1

The rules used in the table are

• a<->b := a&b | !a&!b (for the left)
• a<->b := (a->b) &| (b->a) (for the right)
• Each one of these rules branches into two subgoals, why do you think that there are four? 

Comments

I understood that..I thought a, b|- {} is, path a and path b|-{}.
Thanks for the clarification