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

    ditional exercises 8.1: Applying De Morgans law for quantified statements to logical expressions. Apply De Morgans law to each expression to obtain an equivalent expression in which each negation sign applies directy to a predicate (ie,3x(-Px) v-0(3) is an acceptable final answer, but not-3x P(x) or 3x-(P AQx) (a) -ax P(x) (b) -3x (Px) v Q(x) (c) vx (Px) A Q(x) (a) -x (P(x) A (O) Rx
    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)))

    Expert Answer

     
    a. ⌐∃x P(x) = ∀x ⌐P(x)

    ditional exercises 8.1: Applying De Morgans adjudication coercion quantified statements to argumentative looks. Apply De Morgans adjudication to each look to procure an equipollent look in which each nullifying memorial applies directy to a assert (ie,3x(-Px) v-0(3) is an grateful latest exculpation, still referable-3x P(x) or 3x-(P AQx) (a) -ax P(x) (b) -3x (Px) v Q(x) (c) vx (Px) A Q(x) (a) -x (P(x) A (O) Rx

    Apply De Morgan’s adjudication o each look to procure an equipollent look in which each nullifying memorial applies instantly to a assert (i.e., thereexists x(­P(x) v ­Q(x)) is an grateful latest exculpation, 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 Exculpation

     

    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 apprehend if you keep any demur 🙂