Question

Use De Morgan’s Laws to distribute negations inward (that is, all negations in the resulting equivalent expression should appear before a predicate): ¬∃x(¬∀y(Q(x,y) v ¬R(x,y)) → ∀yP(x,y)). Label each step with the name of the rule used (such as DN: Double Negation, see formula sheet) and which statement numbers the rule uses. The first step is: 1. ¬∃x(¬∀y(Q(x,y) v ¬R(x,y)) → ∀yP(x,y))

Answer #1

- ¬∃x(¬∀y(Q(x,y) v ¬R(x,y)) → ∀yP(x,y))
- ∀x¬(¬∀y(Q(x,y) v ¬R(x,y)) → ∀yP(x,y)) [DeMorgan's]
- ∀x¬(¬¬∀y(Q(x,y) v ¬R(x,y)) v ∀yP(x,y)) [Implication]
- ∀x¬(∀y(Q(x,y) v ¬R(x,y)) v ∀yP(x,y)) [Double negation]
- ∀x(¬∀y(Q(x,y) v ¬R(x,y)) ∧ ¬∀yP(x,y)) [DeMorgan's]
- ∀x(∃y¬(Q(x,y) v ¬R(x,y)) ∧ ¬∀yP(x,y)) [DeMorgan's]
- ∀x(∃y(¬Q(x,y) ∧ ¬¬R(x,y)) ∧ ¬∀yP(x,y)) [DeMorgan's]
- ∀x(∃y(¬Q(x,y) ∧ R(x,y)) ∧ ¬∀yP(x,y)) [Double negation]

* Answer*: ∀x(∃y(¬Q(x,y) ∧ R(x,y)) ∧
¬∀yP(x,y))

*Please up vote. I need it very badly right now.
Comment if you have any doubts. Happy Learning!*

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