# Help proving if $\Sigma\cup\theta\vdash\phi$ then $\Sigma\vdash(\theta\rightarrow\phi)$

For context this is one direction of the Deduction Theorem. And what I am trying to figure out is how the proof in Leary and Kristensen pg. 64, in particular, is supposed to work. They write:

For the more difficult direction we will make use of Proposition 2.2.4. Suppose that $$C=\{\phi : \Sigma\vdash(\theta\rightarrow\phi)\}$$. If we show that $$C$$ contains $$\Sigma\cup\theta$$, $$C$$ contains all the axioms of $$\Lambda$$, and $$C$$ is closed under the rules of inference as noted in Proposition 2.2.4, then by that proposition we will know that $$\{\phi :\Sigma\cup\theta\vdash\phi\}\subseteq C$$. In other words, we will know that if $$\Sigma\cup\theta\vdash\phi$$, then $$\Sigma\vdash(\theta\rightarrow\phi)$$, which is what we need to show.

Proposition 2.2.4 states that $$Thm_\Sigma=\{\phi:\Sigma\vdash\phi\}$$ is the "smallest set" containing all the axioms and closed under the rules of inference.

What I don't understand is how establishing that $$C$$ has these properties proves the direction they say it does. If we suppose $$C=\{\phi : \Sigma\vdash(\theta\rightarrow\phi)\}$$ and then show that $$\{\phi :\Sigma\cup\theta\vdash\phi\}\subseteq C$$, don't we show the opposite direction of what they say we show, namely that if $$\Sigma\vdash(\theta\rightarrow\phi)$$, then $$\Sigma\cup\theta\vdash\phi$$?

I understand their proofs of how $$C$$ bears the properties they set out to prove it has. I just don't get, conceptually, how this shows what they claim it does. Thanks for any help. I really appreciate it.

Long comment

Recall the def of Prop.2.2.4 about $$\text {Thm}_{\Sigma}$$.

In Th.2.7.4 (The Deduction Theorem) the authors prove that:

$$\text {Thm}_{\Sigma \cup \theta} = \{ \phi \mid \Sigma \cup \theta \vdash \phi \} \subseteq \{ \phi \mid \Sigma \vdash (\theta \to \phi) \}=C$$.

If you agree with this, by definition of set inclusion we have that:

if $$\phi \in \text {Thm}_{\Sigma \cup \theta}$$, then $$\phi \in C$$,

that, unwinding the definition of the two sets, means: if $$\Sigma \cup \theta \vdash \phi$$, then $$\Sigma \vdash (\theta \to \phi)$$.

• Oh! That's super interesting/cool! Thanks, Mauro. As usual you're a huge help! Commented May 21, 2021 at 10:22
• @beachtrees - you are welcome :-) Commented May 21, 2021 at 10:23