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For $x,y,z\in\mathbb{R}$, I'm trying to determine the truth value of a particular set of statements

$1) \;\;\;\; \exists x\,|\, \forall y, \exists z\, |\, x+y=z $

$ 2) \;\;\; \exists x\,|\;\forall y\; \text{and}\; \forall z,\, z>y \implies z> x+y $

$3) \;\;\; \forall x, \exists y\; \text{and} \; \exists z \,|\, z>y \implies z>x+y $

$\textbf{My ideas:}$ for $1)$ My reasoning is that it is true since it is always possible to find an $x$ and $z$ such that the statement will remain true.

for $2)$ I believe that it will be false because there does not exists an $x$ such that the relation will hold. For example if a $x$ negative number and $y$ and $z$ are positive and this negative is "large" enough to make the relation false.

for $3)$ I believe it is false for similar reasons as in $2)$.

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    For 2) consider $z > y \to z > 0+y$ – 2017-02-01

1 Answers 1

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1) you're right! You can pick $x=0$ and then for any $y$, pick $z =y$

2) A Mauro says, this is true, since you can pick $x = 0$

3) This statement is true, but for a weird reason: For any $x$, simply pick $y$ and $z$ such that $z \not > y$. Because then the antecedent of the conditional will be false, and hence the whole conditional will be true, and thus the whole statement will be true!