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>
Answer:
6
Step-by-step explanation:
After 10 hours he has graded 10*7 papers which is 70
76-70 = 6
3l yellow paints + 5l blue paints= 8l green paints
To move a graph, f(x), to the right c units, minus c from every x
so f(x) moved to the right 2 units is f(x-2)
A1 is when i = 3. i.e. 2(3) + 3 = 6 + 3 = 9
an is when i = 10. i.e. 2(10) + 3 = 20 + 3 = 23
n = count from 3 to 10 = 8
Therefore, sum = 8/2(9 + 23) = 4(32) = 128
