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## Homework Statement

Consider the following statement:

##\forall x, ((x \in \mathbb{Z} \wedge \neg(\exists y, (y \in \mathbb{Z} \wedge z = 7y))) \rightarrow (\exists z, (z \in \mathbb{Z} \wedge x = 2z)))##

a)Negate this statement.

b)Write the original statement in English.

c) Which statement is true? Explain.

## Homework Equations

## The Attempt at a Solution

a) I negate the statement as ##\forall x, P(x) \rightarrow G(x)##

answer: ##\exists x, ((x \in \mathbb{Z} \wedge \neg(\exists y, (y \in \mathbb{Z} \wedge z = 7y))) \wedge \forall z, (z \in \mathbb{Z} \vee x \ne 2z)##

and then simplifying the statement with y,

= ##\exists x, ((x \in \mathbb{Z} \wedge \forall y, (y \notin \mathbb{Z} \vee z \ne 7y))) \wedge \forall z, (z \notin \mathbb{Z} \vee x \ne 2z)##

b) statement in English: There exists an x, such that x is an integer and for all y, y is an integer or x =/= 2y, and for all z, z is not an integer or x =/= 2z.

c) The negation of the original statement is true.

If we suppose for all x, x is an integer and for all y, y is not an integer or x =/= 2y (the hypothesis in the original statement) then the conclusion, there exists a z such that z is an integer and x = 2z, cannot be true.

**This is because "for all y, y is not an integer or x =/= 2y" <=> "there exists a z such that z is an integer and x = 2z" is a contradiction.**

So the negation of this statement must be true.

My biggest confusion is what i put in bold. I think this might be an unreasonable assumption.. but I don't know how else to see which statement is true. My question is, can I make this assumption because x,y,z seem to be elements of the same universe?

Also I should have named this "quantifier question".. not sure how to edit the title.