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