Negating $\sim(∀x∈ℤ : ∃y∈ℤ: x+y = 5)$

Asked Feb 26 '23 at 10:59

Active Feb 26 '23 at 11:54

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Is the following true?

$$\sim(∀x∈ℤ : ∃y∈ℤ: x+y = 5) \quad≡\quad \sim∀x∈ℤ : ∃y∈ℤ: x+y = 5 ?$$

If not, can somebody explain how i can produce statements that are equivalent to the LHS? I get confused with regards to De Morgan's because I am not used to more than one quantifier and no paranthesis.

edited Feb 26 '23 at 11:54

ryang

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asked Feb 26 '23 at 10:59

Karim Loberg

I suggest avoiding colons and periods in formal sentences. However, if your textbook does use them, then query it to check whether $$∀x: Px\to Q $$ is meant to be read as $$(∀x, Px)\to Q \tag1$$ or as $$∀x, (Px\to Q )\tag2$$ ( $(1)$ and $(2)$ are not equivalent to each other). $$$$ 2. Yes, the LHS and RHS are logically equivalent: logical-operator precedence.

ryang

Feb 26 '23 at 12:03

Intuitively, this statement is saying that every $x$ satisfies some particular property. The negation of that is that there is a counterexample. Symbolically, you should move the negation into the formula step by step. It may help to think of the original formula as $\forall x \in \mathbb{Z}: (\exists y \in \mathbb{Z}: (x+y=5))$ – TomKern Feb 26 '23 at 15:41

Negating $\sim(∀x∈ℤ : ∃y∈ℤ: x+y = 5)$

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