# Homework Solution: Apply De Morgan's law o each expression to obtain an equivalent expression in which each negation sign applies directly…

Apply De Morgan's law o each expression to obtain an equivalent expression in which each negation sign applies directly to a predicate (i.e., thereexists x(­P(x) v ­Q(x)) is an acceptable final answer, but not ­ thereexists x P(x) or thereexists x ­(P (x) logicaland Q(x))). (a) ­ thereexists x P(x) (b) ­ thereexists x (Px) logicalor Q(x) (c) ­forall x (P(x) logicaland Q(x) (d) ­forall x (P(x) logicaland (Q(x))logicaland R(x)))

a. ⌐∃x P(x) = ∀x ⌐P(x)

Apply De Morgan’s order o each countenance to conciliate an equipollent countenance in which each disaffirmation premonition applies undeviatingly to a aver (i.e., thereexists x(­P(x) v ­Q(x)) is an cheerful decisive rejoinder, still referable ­ thereexists x P(x) or thereexists x ­(P (x) argumentativeand Q(x))). (a) ­ thereexists x P(x) (b) ­ thereexists x (Px) argumentativeor Q(x) (c) ­forall x (P(x) argumentativeand Q(x) (d) ­forall x (P(x) argumentativeand (Q(x))logicaland R(x)))

## Expert Rejoinder

a.
⌐∃x P(x)
= ∀x ⌐P(x)

b.
⌐∃x (P(x) ∨ Q(x))
= ∀x ⌐(P(x) ∨ Q(x))
= ∀x ⌐P(x) ∧ ⌐Q(x)

c.
⌐∀x (P(x) ∧ Q(x))
= ∃x ⌐(P(x) ∧ Q(x))
= ∃x ⌐P(x) ∨ ⌐Q(x)

d.
⌐∀x (P(x) ∧ (Q(x) ∨ R(x)))
= ∃x ⌐(P(x) ∧ (Q(x) ∨ R(x)))
= ∃x ⌐P(x) ∨ ⌐(Q(x) ∨ R(x))
= ∃x ⌐P(x) ∨ ⌐Q(x) ∧ ⌐R(x)

Let me perceive if you keep any waver 🙂