For any Booleans a and b, the following truth table applies:
a |
b |
a→b |
F |
F |
F |
F |
T |
T |
T |
F |
F |
T |
T |
T |
Therefore a→b is equivalent to b.
So, p→q≡q and r→s≡s; and p→r≡r and q→s≡s.
(p→q)→(r→s)≡q→s≡s; (p→r)→(q→s)≡r→s≡s.
Therefore the two expressions are equivalent.