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let $P(x,y): x - 7y = 5$ and $Q(x,y): x-3 \geq y + 11$
$\exists x \forall y, Q(x,y)$
$\exists x \exists y, P(x,y) \land \neg Q(x,y)$
$\forall x \exists y, P(x,y) \geq Q(x,y)$

How de we prove if these propositions are true or false, I know when a ($\forall x, P$) proposition is false you can prove it by showing that ($\exists x, \neg P$) is true, and to prove ($\forall x, P$) is true you have to use a detailed demonstration, and to prove ($\exists x, P$) you just show one example where the equation is true. However in these cases I really have no idea how to prove such propositions.

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  • $\begingroup$ Is there a typo: $Q(x,y)$ does not contain the variable $y$ in its specification? $\endgroup$
    – user2661923
    Commented Feb 11, 2021 at 5:52
  • $\begingroup$ @user2661923 yes sorry I just corrected it Thank you! $\endgroup$
    – William
    Commented Feb 11, 2021 at 6:03
  • $\begingroup$ @Stacker you converted them to their negation form but I don't see how this answer my question (don't mean to be ungrateful), is there something missing? $\endgroup$
    – William
    Commented Feb 11, 2021 at 6:05
  • $\begingroup$ Sorry, I misunderstood your question. For the first one you find an x such that for all y, Q(x,y) holds. So one x where Q(x,y) is true for ALL y. No such x exists, so the first statement is false. $\endgroup$
    – Vons
    Commented Feb 11, 2021 at 6:10
  • $\begingroup$ @Stacker Yep that's exactly what I thought but how de we prove that $\endgroup$
    – William
    Commented Feb 11, 2021 at 6:16

1 Answer 1

Reset to default
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To disprove the first one set y = x.
To determine the second one, use the equation in P to reduce Q to a single variable and make your decision.
Explain why the third one is nonsense.

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  • $\begingroup$ Thank you but what do you mean by "use the equation in P to reduce Q to a single variable" I don't get it $\endgroup$
    – William
    Commented Feb 11, 2021 at 7:35
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    $\begingroup$ @Dom. P is an equation. Solve for x in terms of y. Use that value for x in Q to get an inequality in y. $\endgroup$
    – William Elliot
    Commented Feb 11, 2021 at 9:05
  • $\begingroup$ Is that a proper explanation why it's true? We have $x - 7y = 5$ and $x - 3 < y + 11$. Therefore $y = (x-5) / 7$ from $P$ \begin{align*} &x-3 < (x-5) / 7 +11 \\ &7x -21 < x + 72 \\ &6x < 93 \\ & x < 93 / 6 \\ \end{align*} Suppose $x = 5$. Therefore $y = 0 / 7 \Leftrightarrow y = 0$. We conclude the porposition is True because such $x$ and $y$ exists $\endgroup$
    – William
    Commented Feb 11, 2021 at 17:55

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