I had to negate the following expression:
$\underset{A\subseteq\mathbb{N}}\forall \, \underset{m\in\mathbb{N}}{\exists!} [m\in A \land \underset{a\in A}\forall m\leq a]$

Surely I expanded the quantifier and worked it out step for step and got:
$\underset{A\subseteq\mathbb{N}}\exists \, \underset{m\in\mathbb{N}}\forall [(m\in A \land \underset{a\in A}\forall m\leq a) \Rightarrow \underset{n\in\mathbb{N}}\exists((n\in A \land \underset{a\in A}\forall n\leq a) \land m\neq n)]$

So my question is if this is correct? I used 1.21 (p. 18) of this lecutre http://www.math.lmu.de/~philip/publications/lectureNotes/calc1_forInfAndStatStudents_new.pdf

The correction of my solution said that the antecedent was wrong, but I don't really get it why it should be wrong as per definition in the lecture notes.


You got that negation absolutely correct!

To see this intuitively: the original sentence says that in every subset of natural numbers there is exactly one number that's smaller or equal to every number in that subset. Roughly put: every subset of natural numbers has exactly one 'smallest' natural number.

So, if you negate that, then there is some subset of natural numbers so that either there is no 'smallest' number in that subset or there are at least two 'smallest' numbers in that subset ... which can be captured by saying if there is a smallest number in that subset, then there is a second (i.e. different) smallest number in that subset. And that's exactly what your statement is saying.


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