1
$\begingroup$

I can show that $\forall x \phi \vdash \exists x \phi$ through a direct deduction as follows, using axioms as defined in Enderton

  1. $(\forall x) \phi$ by hypothesis.

  2. $ (\forall x) \phi \rightarrow \phi$, Axiom.

  3. $ \phi$, modus ponens on lines 1 and 2.

  4. $(\forall x)\neg \phi \rightarrow \neg \phi$, Axiom.

    1. $ ((\forall x) \neg \phi \rightarrow \neg \phi )\rightarrow ( \phi \rightarrow (\exists x) \phi)$, Axiom (tautology).

    2. $ \phi \rightarrow (\exists x) \phi$, modus ponens on lines 4 and 5.

    3. $(\exists x) \phi$, modus ponens on lines 3 and 6.

However, at this point, I now need to invoke the deduction theorem in order to conclude that $\vdash \forall x \phi \rightarrow \exists x \phi$. I want to avoid metalogical results and be purely syntactic. I know that you can prove the deduction theorem through a direct deduction, but I'm wondering how I can adjust my specific proof directly to get the result through just axioms and modus ponens.

I think that if I can show $\vdash \forall \phi \rightarrow ( \phi \rightarrow (\exists x) \phi)$, I could apply the $T$ rule. But I'm not sure. Help appreciated.

  • 1
    Note that the deduction theorem for which you have a proof is not really a theorem in that system (it's a theorem outside the system). To be strict the theorem says that given $\psi\vdash\phi$ you can prove $\psi\rightarrow\phi$ and the proof for it concretely shows how you construct the proof. The theorem is used to just get from here to there without repeating the steps, but one could say that this is just a shorthand for producing the steps in the proof of the deduction theorem. – 2017-02-17
  • 0
    The *proof* of the deduction theorem is actually an algorithm: it shows you how to take a formal derivation using the deduction theorem and turn it into one that does not. So you could simply follow that algorithm to convert your derivation into a different one. – 2017-02-17

1 Answers 1

2

The "trick" to get rid of the Deduction Th is to prefix (in a suitable way) the assumption that you want to "discharge" with DT to the steps in the derivation; see Stephen Cole Kleene, Mathematical logic, Dover (1967), page 40.

But in the present case, the needed derivation is quite straightforward:

1) $\vdash (∀x)ϕ→ϕ$ , Axiom.

2) $\vdash (∀x)¬ϕ→¬ϕ$ , Axiom.

3) $\vdash (∀x)¬ϕ→¬ϕ)→(ϕ→(∃x)ϕ)$ , Axiom (tautology).

4) $\vdash ϕ→(∃x)ϕ$ , modus ponens on lines 2 and 3.

With a suitable instance of the Axiom (tautology) :

5) $\vdash (p \to q) \to ((q \to r) \to (p \to r))$

6) $\vdash (∀x)ϕ → (∃x)ϕ$ , modus ponens twice on lines 1, 4 and 5.

Note

In Enderton's system this formula is valid because the semantics for FOL used is the "traditional" one, with nonempty domain $D$.

This is consistent with the fact that in Free Logic and Inclusive Logic, i.e. logics cosistent with the assumption that $D$ may be empty, the schema:

$∀xA → A$,

is not valid