Having trouble proving a statement about the consequences of two sets

Question

The following is a problem from a course in Logic I'm attending:

Let $\Gamma_1$ and $\Gamma_2$ be two sets of formulae. Prove that $$\text{Cons}(\Gamma_1 \cup \Gamma_2) = \text{Cons}(\Gamma_1) \cup \text{Cons}(\Gamma_2) \Rightarrow \Gamma_1 \subseteq \text{Cons}(\Gamma_2) \lor \Gamma_2 \subseteq \text{Cons}(\Gamma_1)$$

$\text{Cons}(\Gamma)$ is the set of consequences of $\Gamma$

The problem actually asks for the converse as well but I had no trouble proving that. Here, however, I always end up getting stuck, trying to do it directly, by contra-reciprocal or by absurd.

Any tip on how to prove this?

Answer 1

Assume not, i.e. :

$Γ_1 \nsubseteq \text {Cons}(Γ_2)$ and $Γ_2 \nsubseteq \text {Cons}(Γ_1)$.

This means that there are formulas $\varphi_1, \varphi_2$ such that :

$\varphi_1 \in \Gamma_1 \text { and } \varphi_1 \notin \text {Cons}(Γ_2)$

and

$\varphi_2 \in \Gamma_2 \text { and } \varphi_2 \notin \text {Cons}(Γ_1)$.

But obviously $\Gamma_i \subseteq \text {Cons}(Γ_i)$, and thus from $\varphi_i \in \Gamma_i$ we have that $\varphi_i \in \text{Cons}(Γ_1 \cup Γ_2)$.

So, $\varphi_1,\varphi_2 \in \text{Cons}(Γ_1 \cup Γ_2)$ and finally, $\varphi_1 \land \varphi_2 \in \text{Cons}(Γ_1 \cup Γ_2)$.

But from the fact that $\varphi_1 \notin \text {Cons}(Γ_2)$ and $\varphi_2 \notin \text {Cons}(Γ_1)$ it follows that $\varphi_1 \land \varphi_2 \notin \text{Cons}(Γ_1)$ and $\varphi_1 \land \varphi_2 \notin \text{Cons}(Γ_2)$ (because $\varphi_1 \land \varphi_2 \vDash \varphi_i$).

So $\varphi_1 \land \varphi_2 \notin \text{Cons}(Γ_1) \cup \text{Cons}(Γ_2)$, contradicting the fact that :

$\text{Cons}(Γ_1 \cup Γ_2) = \text{Cons}(Γ_1) \cup \text{Cons}(Γ_2)$