$X_\rightarrow =\{(a\rightarrow b)|\alpha \in \sum_1,\beta \in \sum_2\}$ Proof: $\sum_1 \cup \sum_2 \models X_{\rightarrow}$

Question

I have the following question :

$\sum_1,\sum_2 \subseteq WFF$

$X_\rightarrow =\{(a\rightarrow b)|\alpha \in \sum_1,\beta \in \sum_2\}$

Proof: $\sum_1 \cup \sum_2 \models X_{\rightarrow}$

I'm not sure how to approach this since $\sum_1 \cup \sum_2$ is a group of atoms but $X_{\rightarrow}$ using two different atoms.

For example : Assuming $\sum_1 \cup \sum_2=\{p_0,p_1\}$ and lets assume $z \models \sum_1 \cup \sum_2$ so $z(p_0)=T,z(p_1)=T$ but $X_{\rightarrow}$ has two parameters? so how to use $z$? since $z$ gets one parameter yet.

I'm confused any ideas?

Answer 1

I'm not sure how to approach this since $\Sigma_1 \cup \Sigma_2$ is a group of atoms but $X_{\rightarrow}$ using two different atoms.

It's not necessarily true that $\Sigma_1 \cup \Sigma_2$ is a set of atoms. The given is that $\Sigma_1 \cup \Sigma_2$ is a subset of the set of well-formed formulas (denoted by $WWF$).

You want to prove that $\Sigma_1 \cup \Sigma_2\models \{(\alpha\rightarrow \beta)|\alpha \in \Sigma_1\land \beta \in \Sigma_2\}$. By definition of $\models$ the thesis is that for all valuations $v$, $$\forall \delta \in \Sigma_1 \cup \Sigma_2(v(\delta)=T)\implies\forall \gamma \in \left\{(\alpha\rightarrow \beta)|\alpha \in \Sigma_1\land \beta \in \Sigma_2\right\}(v(\gamma)=T) \tag 1$$

So prove this statement.

Assume the antecedent and try to prove that $\forall \gamma \in \left\{(\alpha\rightarrow \beta)|\alpha \in \Sigma_1\land \beta \in \Sigma_2\right\}(v(\gamma)=T)$.
To do this take an arbitrary $\gamma$ in $X_\to$. By definition of $X_\to$, there exist $\alpha$ in $\Sigma_1$ and $\beta$ in $\Sigma_2$ such that $\gamma=\alpha\to \beta$.

Recall that you want to prove that $v(\gamma)=T$. To get to this use the antecedent of $(1)$.