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$\begingroup$

Is the negation of:

$ x < y∧(∃z)[x < z < y] $

equal to:

$ x ≥ y \vee (∀z)[(x ≥ z) \vee (z ≥ y)] $

? I am confused on the negation of x < z < y

2 Answers 2

1

$x

hence, its negation is $ x \ge z \lor z \ge y$

So yes, you did everything corectly!

  • 0
    thanks! is the negation of the whole statement correct? – 2017-02-12
  • 0
    @user7252850 yeah, the rest was all fine! – 2017-02-12
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We read ($\ref{1}$) as ($\ref{2}$)

$x

$(x

so its negation is ($\ref{3}$) as you have identified in your expression.

$(x\geq y) \lor (y\geq z)\label{3}\tag{3}$

The full worked negation is:

$\lnot(x

$\lnot(x

$(x\geq y) \land (\forall z)[(x\geq y) \lor (y\geq z)])$

So your work is indeed correct.