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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?

Git Gud
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skyfire
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1 Answers1

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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.

Erik M
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  • It would be better if you typed out a little bit of what's in the link. – qwr Mar 19 '14 at 04:11
  • 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 – skyfire Mar 19 '14 at 04:39
  • 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 $z>x+y$. – Erik M Mar 19 '14 at 04:46
  • Its question 2.5(f) (from 4th edition) – skyfire Mar 19 '14 at 04:47
  • I have edited my answer above. – Erik M Mar 19 '14 at 04:53
  • Sorry can you explain what the "p" part of the statement is considered false. I think I am starting to understand what is going on, but im a little confused if the problem stems from the "p" part being wrong from how the statement was quantified or if it was because we were able to violate the z>y because we were able to choose z – skyfire Mar 19 '14 at 04:59
  • I'm not sure if I understand the difference between the two alternatives you provided, we were able to choose $z\le y$ because of the quantifiers, making the statement $z>y$ false. – Erik M Mar 19 '14 at 05:04
  • @ErikMiehling You're wrong. You don't interpret the statement correctly, because you don't give importance to the priority of quantifiers over the implication. There is also a $\forall x$ that has to be used. – frabala Mar 19 '14 at 05:04
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    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. – Erik M Mar 19 '14 at 05:32
  • You're right @ErikMiehling, sorry. – frabala Mar 19 '14 at 06:12
  • No problem, the discussion helped clear up some of the loose ends. – Erik M Mar 19 '14 at 07:21
  • I know this is an old question, but I do not get a lot of time to work on this stuff and have recently come back to this. And really the only part of confusion I have left is -what rule of inference allowed you to assume that z<=y. Existential instantiation does not allow you to instantiate a variable...and I believe it also requires an introduction of a brand New symbol that does not appear in the premises or conclusion. So what gives? – skyfire Jan 27 '15 at 21:42