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Let $\mathcal{L}_0=\mathcal{L}[\{\neg, \rightarrow\}]$. Define the system $L_0$ as follows:

An axiom of $L_0$ is any formula of $\mathcal{L}_0$ of the form

  • (A1) $(\alpha \rightarrow ( \beta \rightarrow \alpha))$
  • (A2) $((\alpha \rightarrow ( \beta \rightarrow \gamma) \rightarrow ((\alpha \rightarrow \beta) \rightarrow ( \beta \rightarrow \gamma)))$
  • (A3) $((\neg \beta \rightarrow \neg \alpha ) \rightarrow (\alpha \rightarrow \beta))$

The only rule of inference in $L_0$ is modus ponens, i.e. from $\alpha$ and $(\alpha \rightarrow \beta)$ infer $\beta$.

A formula of $\mathcal{L}_0$, $\phi$, is provable from a set of formulas $\Gamma$ in the system $SQ$, if there exists a finite sequence of sequents where each follows from the next according to the following set of rules:

  • (Ass) If $\psi \in \Delta$, we infer $\Delta \vdash_{SQ} \psi$
  • (MP) If $\Delta \vdash_{SQ} \psi$ and $\Delta ' \vdash_{SQ} (\psi \rightarrow \phi)$ then infer $\Delta \cup \Delta ' \vdash_{SQ} \phi$
  • (DT) If $\Delta \cup \{\psi \} \vdash_{SQ} \chi$ we infer $\Delta \vdash_{SQ} (\psi \rightarrow \chi)$
  • (PC) If $\Delta \cup \{ \neg \psi \} \vdash_{SQ} \chi$ and $\Delta ' \cup \{ \neg \psi \} \vdash_{SQ} \neg \chi$ infer that $\Delta \cup \Delta ' \vdash_{SQ} \psi$

Prove that the systems $SQ$ and $L_0$ are equivalent, i.e. that $\Gamma \vdash_{L_0} \phi$ iff $\Gamma \vdash_{SQ} \phi$.

Ideally the legwork for this has been done elsewhere, in which case a link will be gratefully received.

Is the idea then to show that $\emptyset \vdash_{SQ} (A1)$, $\emptyset \vdash_{SQ} (A2)$, $\emptyset \vdash_{SQ} (A3)$ and that $SQ$ somehow implies (MP) in $L_0$, and then to do the reverse, i.e. show that $L_0$ proves each of the sequents of $SQ$? I don't quite see how that would be formalised.

Mathmo
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  • Basically, you are on the right track; in $SQ$ you have the rule (MP) that is already modus ponens; thus, it is enough to "rewrite" (MP) with $\Delta = \Delta' = \emptyset$. For the other "direction", you need Deduction Theorem (provable from (A1)-(A3) with the help of $\vdash \alpha \rightarrow \alpha$, which is easily derivable from them), i.e. from $\Gamma \cup \alpha \vdash \beta$, infer $\Gamma \vdash (\alpha \rightarrow \beta)$: this will play the role of (DT). About (PC), it is "simply" (A3) in sequent-form. – Mauro ALLEGRANZA Apr 15 '14 at 13:45
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    You have to "consult" a textbook which use the same $L_0$ axiom system, like Elliott Mendelson, Introduction to Mathematical Logic (4th ed - 1997) : page 35, with "deduction from $\Gamma$ and Deduction Theorem (page 37). They are the "ingredients" for deriving in it (Ass) (trivial) and (MP) (see comment above) and (DT) and (PC) (as above). I suggest you also : Mordechai Ben-Ari, Mathematical Logic for Computer Science (3rd ed - 2012) : Chapter 3 (page 49-on); the "mechanism" is explained but, unfortunately, the sequent calculus system is different... – Mauro ALLEGRANZA Apr 15 '14 at 14:19
  • See Wiki for an on-line resource regarding how to prove Deduction Theorem with your $L_0$. – Mauro ALLEGRANZA Apr 15 '14 at 14:59

1 Answers1

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I suppose there is a mistake; your (A2) must be :

$(α→(β→γ)) \rightarrow ((α→β)→(α→γ))$.

From $SQ$ to $\mathsf L_0$ we have :

(A1)

1) --- $\alpha, \beta \vdash \alpha$ --- by (Ass)

2) --- $\alpha \vdash \beta \rightarrow \alpha$ --- from 1) by (DT)

3) --- $\vdash \alpha \rightarrow (\beta \rightarrow \alpha)$ --- from 2) by (DT).

(A2)

1) --- $\alpha \rightarrow (\beta \rightarrow \gamma) \vdash \alpha \rightarrow (\beta \rightarrow \gamma)$ --- by (Ass)

2) --- $\alpha \vdash \alpha $ --- by (Ass)

3) --- $\alpha, \alpha \rightarrow (\beta \rightarrow \gamma) \vdash \beta \rightarrow \gamma$ --- from 1), 2) by (MP)

4) --- $\alpha \rightarrow \beta \vdash \alpha \rightarrow \beta $ --- by (Ass)

5) --- $\alpha, \alpha \rightarrow \beta \vdash \beta $ --- from 2), 4) by (MP)

6) --- $\alpha, \alpha \rightarrow \beta, \alpha \rightarrow (\beta \rightarrow \gamma) \vdash \gamma$ --- from 3), 5) by (MP)

7) --- $\alpha \rightarrow \beta, \alpha \rightarrow (\beta \rightarrow \gamma) \vdash \alpha \rightarrow \gamma$ --- from 6) by (DT)

8) --- $\alpha \rightarrow (\beta \rightarrow \gamma) \vdash (\alpha \rightarrow \beta) \rightarrow (\alpha \rightarrow \gamma)$ --- from 7) by (DT)

9) --- $\vdash (\alpha \rightarrow (\beta \rightarrow \gamma)) \rightarrow ((\alpha \rightarrow \beta) \rightarrow (\alpha \rightarrow \gamma))$ --- from 8) by (DT).

(A3)

1) --- $\lnot \beta \rightarrow \lnot \alpha \vdash \lnot \beta \rightarrow \lnot \alpha$ --- by (Ass)

2) --- $\lnot \beta \vdash \lnot \beta$ --- by (Ass)

3) --- $\lnot \beta, \lnot \beta \rightarrow \lnot \alpha \vdash \lnot \alpha$ --- from 1), 2) by (MP)

4) --- $\alpha, \lnot \beta \vdash \alpha$ --- by (Ass)

5) --- $\alpha, \lnot \beta \rightarrow \lnot \alpha \vdash \beta$ --- from 4), 5) by (PC)

6) --- $\lnot \beta \rightarrow \lnot \alpha \vdash \alpha \rightarrow \beta$ --- from 5) by (DT)

7) --- $\vdash (\lnot \beta \rightarrow \lnot \alpha) \rightarrow (\alpha \rightarrow \beta$) --- from 6) by (DT).

The only rule of inference of $\mathsf L_0$ is modus ponens; we derive it from (MP) simply with $\Delta = \Delta' = \emptyset$.

Mauro ALLEGRANZA
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