Contrapositive fails:

A set A is infinite if it cannot be decomposed into finite sets using a finite number of bipartitions

-> If it can be decomposed into infinite sets using a infinite number of bipartitions, then a set A is finite

@Secret I hope you're joking, but if you're not then you're making a very basic logical error, again.

Well I knew an error is made because the conclusion is clearly absurd, thus the "fail" means [attempt to find a] contrapositive fails

though I still need more revision on classical logic as clearly I am still very weak on passing negations inside quantifiers

@Secret Rewrite the two versions you have above in logical form, and we can proceed from there.

not (finite disjoint union of finite sets) $\implies$ A is not finite

not $(\forall n \in \Bbb{N} [A=\sqcup_n A_n$ and $|A_n|\in \Bbb{N}]) \implies |A|\not\in \Bbb{N}$

->

$|A| \in \Bbb{N} \implies \forall n \in \Bbb{N} [A = \sqcup_n A_n$ and $|A_n|\in \Bbb{N}]$

hmm it makes sense now

Looks like the not never made into the quantifier, thus it got cancelled out when the contrapositive is taken

Meanwhile, there seemed to be something interesting when we expand the not () in the original statement:

not $(\forall n \in \Bbb{N} [A=\sqcup_n A_n$ and $|A_n|\in \Bbb{N}])

--> $\exists n \in \Bbb{N} [A \neq \sqcup_n A_n$ or $|A_n| \not\in \Bbb{N}]$

by de morgan

The latter half of that makes sense since if any element in the disjoint union is infinite, then the set has to be infinite. However, that $A$ cannot be decomposed into a certain finite disjoint union of finite sets $A_n$ is sufficient to deduce $A$ infinite seemed to be not true, as e.g. the singleton cannot be decomposed into any disjoint union of m sets where $m > 1$

and it is clearly finite

so that means, there is something I forgot after passing the negation into the universal quantifier (which changes it into a existential quantifier)

@Secret Your first sentence is correct. Your mistake was simply in putting "not" everywhere, which of course does not work.

@Secret But your second sentence is wrong, though at least you didn't make the same negation error.

I suggest you start with more basic arithmetic first. Express "k is the sum of three squares" using a first-order arithmetic statement.

Then you will see what I mean.

$\exists a \in \Bbb{N}[\exists b \in \Bbb{N}[\exists c \in \Bbb{N}[k=a^2+b^2+c^2]]]$

@Secret Right. Similarly, to say "A is a finite union of finite sets" you need to say "∃F ( F is finite ∧ every member of F is finite )".

You cannot just introduce notation such as "A_n" with no declaration of what it means, and in this case it's unnecessary and cumbersome to introduce sequences when we can state what we want directly. You just need to use the definition of "finite" to expand "X is finite".

Sorry I forgot about A... "∃F ( A = Union(F) ∧ F is finite ∧ every member of F is finite )".