 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2005/calculus-linear-algebra-discrete-mathematics-logic-and-proof-and-models-of-computation/8/d/solution

The first two clauses now require

x = 2 z

and

y = 2 z

. Thus they already imply

x = y

. Choose, for example,

x = 1, y = 2

z

can satisfy both equations. Consequently the existential quantifier fails for this pair, irrespective of the universally quantified final clause, and

N^{''} \neq ⊨ ψ .

In fact even

x = y = 1

fails, since

1 = 2 z

has no solution among the natural numbers.

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