Question

i.
The negation of p ^(q r)
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Answer 1

$\huge \sf {\orange {\underline {\pink{\underline {A᭄ɴsᴡᴇʀ࿐ :−}}}}}$

$\pink{▬▬▬} \red{▬▬▬} \green{▬▬▬} \blue{▬▬▬} \orange{▬▬▬}$

By definition, p → q is false if, and only if, its hypothesis, p, is true and its conclusion, q, is false. It follows that the negation of "If p then q" is logically equivalent to "p and not q."

p→q =∼p∨q−−−(1)

∼[(p→q)∧r]

∼[∼p∨q∧r]...[form(i)]

∼(∼p∨q)∨∼r...[Demorgans low]

∼(∼p)∧∼q∨q∼r av(b∧c)

p∧∼q∨∼r =(a∨b)∨(a∨c)]

$\pink{▬▬▬} \red{▬▬▬} \green{▬▬▬} \blue{▬▬▬} \orange{▬▬▬}$