By Direct Proof :
1.(∼H(x)∨∼S(x))→(P(x)∨L(x))1.(∼H(x)∨∼S(x))→(P(x)∨L(x))
2.P(x)→E(x)2.P(x)→E(x)
3.∼E(x)3.∼E(x)
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
4.H(x)4.H(x)
5.P(x)→E(x)≡∼P(x)∨E(x)5.P(x)→E(x)≡∼P(x)∨E(x) by Material Implication
6.∼P(x)6.∼P(x) , #5 and #3 by Disjunctive Syllogism
7.∼P(x)∨∼L(x)7.∼P(x)∨∼L(x) , #6 by Addition ( I just add ∼∼L(x))
Since #7 is logically equivalent to ∼(P(x)∨L(x))∼(P(x)∨L(x)) by De Morgan's Law,
8.∼(∼H(x)∨∼S(x))8.∼(∼H(x)∨∼S(x)) , #1 and #7 by Modus Tollens.
Distributing the ∼∼, we'll have,
9.H(x)∧S(x)9.H(x)∧S(x) by De Morgan's and Double Negation
10.H(x)10.H(x) by Simplification ■