³x.Vy.Vz.F(x, y, z) = \x.]y.Vz.¬F(x, y, z)

C++ for Engineers and Scientists
4th Edition
ISBN:9781133187844
Author:Bronson, Gary J.
Publisher:Bronson, Gary J.
Chapter4: Selection Structures
Section: Chapter Questions
Problem 14PP
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Show that this is true or else give a counterexample. 

3x. Vy. Vz.F(x, y, z) = Vx.y.Vz.¬F(x, y, z)
Transcribed Image Text:3x. Vy. Vz.F(x, y, z) = Vx.y.Vz.¬F(x, y, z)
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I'm not quite understanding how the right-hand side is not holding up. In the example given, if we fix x at any value, independent of the value y takes, wouldn't there be an equivalence for one value of z? For example, if x=1, then if we choose any value of y we can say 1+y = z. For z = y-1 (and only this value), this equality would be valid and therefore the predicate, ∃x.∀y.∀z x + y ≠ z would be false as there is a value of z that completes the equality going against the for all z quantifier.

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