Move the negation inside as far as it will go:

~(∃x∈S)(∀z∈T)( ∃y∈U) x ≥ sin z ⇒ y

Question

Move the negation inside as far as it will go:

~(∃x∈S)(∀z∈T)( ∃y∈U) x ≥ sin z ⇒ y < x^2 + z^2

terenzreignz · Answer

You know how to bring a negation inside a quantifier?

terenzreignz · Answer

Keep three things in mind here:
To bring a negation inside an existence quantifier, change it to a universal quantifier
and then negate the inside.
To bring a negation inside a universal quantifier, change it to an existence quantifier and then negate the inside.
To negate a statement of the form p => q, it would become p ^ ~q (p and not q)

terenzreignz · Answer

So here goes... $$\sim(\exists x \in S)(\forall z \in T)(\exists y \in U)\ x \ge \sin z \rightarrow y < x^{2} + z^{2}$$First, bring the negation inside the first existence quantifier... $$(\forall x \in S)\sim(\forall z \in T)(\exists y \in U)\ x \ge \sin z \rightarrow y < x^{2} + z^{2}$$Next, bring the negation inside the universal quantifier... $$(\forall x \in S)(\exists z \in T)\sim(\exists y \in U)\ x \ge \sin z \rightarrow y < x^{2} + z^{2}$$Then, bring the negation inside the second existence quantifier... $$(\forall x \in S)(\exists z \in T)(\forall y \in U)\sim (x \ge \sin z \rightarrow y < x^{2} + z^{2})$$ Note how the quantifier changed everytime the negation crossed it. Now it is down to negating something of the form p => q, which we know now to be p ^ ~q (Since p -> q is equivalent to ~p v q, and the negation of ~p v q is p ^ ~q) Therefore, it is down to... $$(\forall x \in S)(\exists z \in T)(\forall y \in U)x \ge \sin z\ AND \ y \ge x^{2} + z^{2}$$ (Remember that the negation of less than is greater than OR equal to...) :D

anonymous · Answer

Thanks guys, I just didn't know whether I should keep going after bringing the negation inside the quantifiers :)