# How to get around circular reasoning in the definitions of the logical conjunction and the cartesian product

I have been thinking about how one might build up mathematics from scratch. After a bit of thought I realised that I would have to develop logic, set theory, and functions before I could even begin to talk about what a number is.

I have defined the logical conjunction as a binary operator $\wedge: \{T, F\}^2 \to \{T, F\}$ satisfying $$T \wedge T = T, \ \ T \wedge F = F, \ \ F \wedge T = T, \ \ F \wedge F = F,$$ and I have defined the cartesian product as $$A \times B = \{(a, b) \ | \ a \in A \ \wedge \ b \in B\}.$$

However there is some circular reasoning here; the domain of the logical conjunction is $\{ T, F \}^2$ so it depends on the definition of the cartesian product, but the cartesian product depends itself on the logical conjunction.

I have already defined everything else I have used here (i.e. 'set', 'element', 'set membership', 'function', 'domain', 'codomain', 'ordered pair', 'variable', and 'predicate'), the only problem is with $\wedge$ and $\times$.

Can anyone think of a way of solving this problem? Perhaps defining $\wedge$ as a binary operator is overkill, and this is the cause of the problem, but if one wants to do everything proper, surely this is the right way to define it?

(P.S. please don't just say that my question is moot because of incompleteness and ZFC and so on; I'm just having some fun seeing how far I get can get building up mathematics.)

• Oct 16, 2017 at 19:53
• Logic is the foundation upon which set theory is built. The conjuction of statements is defined in logic. So to start from scratch, first build or assume a logic, perferably a FOL. Oct 16, 2017 at 20:10
• You don't want to make T and F objects in your theory, nor treat logical operators as functions. If you do, Russell's paradox will destroy it. Leave the logic to metamathematics. You might want to examine how formalisms introduce logic and proof. Personally, I've found Bourbaki's Elements of Set Theory enlightening. Oct 16, 2017 at 23:44
• Update: I asked one of my lecturers about this and they came up with the following solution. Oct 18, 2017 at 17:03
• First, for an element $a$ and a set $B$, define $$a \otimes B = \{ (a, b) \ | \ b \in B \}.$$ Then define $$A \times B = \{ a \otimes b \ | \ a \in A \}$$ for sets $A$ and $B$. However then I realised that the catesian product is an operator from sets to sets. So if you denote the 'infinite set of all sets', whatever that supposedly means, by $\mathbb{S}$ - or even if choose some more agreeable domain $D$ - then we have $\times: D^2 \to D$, but $D^2 = D \times D$, so the cartesian product appears in its own definition! Oct 18, 2017 at 17:09