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Hi I am teaching myself analysis and bought "Analysis - With an introduction to Proof" by Steven R. Lay. Now one of the practice problems is "Determine the truth value of each statement, assuming x, y and z are real numbers" $$ \forall x, \exists y \, and \, \exists z \, such\, that\, z>y\, implies\, that\, z>x+y $$

Now the solution manual says "True. Take z<=y. This makes "z>y" false so that the implication is true." Now I find this "solution" troubling because I thought the whole idea behind proving an implication was to assume the antecedent IS true. Is this a mistake by the author?

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It is not a mistake; see the truth table here.

The main idea is that in the logical implication $p\Rightarrow q$, if $p$ is false, then the implication does not actually promise anything about $q$. My math prof at the time told the class: irrelevance does not imply falsehood.

In your problem, $z$ and $y$ are preceded by existential quantifiers, so we can choose $z$ and $y$, and thus we choose $z$ and $y$ such that $z\le y$, making the implication true.

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  • $\begingroup$ It would be better if you typed out a little bit of what's in the link. $\endgroup$
    – qwr
    Commented Mar 19, 2014 at 4:11
  • $\begingroup$ I don't understand what makes the "p" part false in this example. What is wrong with saying there exists a pair (y,z) such that if z>y then z>x+y $\endgroup$
    – skyfire
    Commented Mar 19, 2014 at 4:39
  • $\begingroup$ Are you sure you have the question correct? I am looking at Lay 2.6(f) and it asks $\forall x$ and $\forall y$, $\exists z$ such that $z>y$ implies that $z>x+y$. $\endgroup$
    – Erik M
    Commented Mar 19, 2014 at 4:46
  • $\begingroup$ Its question 2.5(f) (from 4th edition) $\endgroup$
    – skyfire
    Commented Mar 19, 2014 at 4:47
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    $\begingroup$ Recall that the universal quantifier can be distributed across an implication, that is, $\forall x (P\Rightarrow Q) \equiv \forall x P \Rightarrow \forall x Q$. Using this, we can see that the original problem can equivalently be written as $(\forall x \exists y\exists z \text{ such that } z>y) \Rightarrow (\forall x\, z>x+y)$. Now, following my answer above, we can choose $z$ and $y$ to make the implication true. $\endgroup$
    – Erik M
    Commented Mar 19, 2014 at 5:32

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