1

I am reading "Mathematical logic" by Ian chriswell and Hodges and at one point in the text they mention the deductive theorem (page 17) which states;

If $\Gamma \cup \left \{ \phi \right \} \vdash \psi $ then $\Gamma \vdash (\phi \rightarrow \psi) \quad$ where $\Gamma$ is the set containing the undischarged assumptions of the argument. Now my confusion starts at the point where they mention that its possible that $\phi \in \Gamma$ in which case it will still be an assumption in $\Gamma \vdash (\phi \rightarrow \psi) \quad$ however I was under the impression that the whole point of the deduction theorem was to enable us to discharge assumptions so that they no longer are apart of $\Gamma$, so what exactly is going on in this particular case? Is $\phi$ discharged or not?

The author also mentions that the rule is still correctly applied if we do not discharge all occurrences of $\phi$ which does not seem like a correct statement for a rule whose purpose is to discharge assumptions.

Git Gud
  • 31,356
skyfire
  • 815

1 Answers1

4

We can discharge the assumption $\phi$, but we are not forced to do so.

See page 17 :

Thus if $\phi$ is an assumption written somewhere in $D$, then we discharge $\phi$ by writing a "dandah" through it: $\phi^/$. In the rule ($\to$I) below, and in similar rules later, the $\phi$ means that in forming the derivation we are allowed to discharge [emphasis mine] any occurrences of the assumption $\phi$ written in $D$.

The rule is still correctly applied if we do not discharge all of them [emphasis mine]; in fact the rule is correctly applied even if $\phi$ is not an assumption of $D$ at all, so that there is nothing to discharge.

You have also to note that [page 17] :

We shall discharge occurrences of assumptions.

Again, we can discharge all occurrences of an assumption $\phi$ but we are not forced to do it: we can discharge some occurrences of $\phi$ but not all. In this way, after the "discharge" the set $\Gamma$ of assumption may still include occurrences of $\phi$.

The "idea" of an undischarged assumption is simply this : if we can prove $\vdash \varphi \to \psi$ (without assumption) we can also prove : $\sigma \vdash \varphi \to \psi$ (with an extra-assumption).

Intuitively, a "redundant" assumption does not invalidate the proof of a theorem.

Mauro ALLEGRANZA
  • 94,169
  • Is there a particular reason why you would not want to discharge all occurrences? – skyfire Mar 08 '15 at 15:36
  • @skyfire - it can be useful : see Example 2.4.4 page 9. – Mauro ALLEGRANZA Mar 08 '15 at 15:44
  • I think I was considering your explanation but it didn't seem right to me because I'm afraid that I would have to worry about the scope of the assumption if I left some occurrences undischarged and others not discharged – skyfire Mar 08 '15 at 15:45
  • Yeah I looked at the example but there ϕ was still completely discharged and not apart of the undischarged set $\Gamma$, maybe by discharging occurrences he meant every ϕ in D gets discharged but the step at which it becomes discharged may happen later? For example in 2.4.4 the first step of application is left discharged but ϕ still eventually gets charged later on? – skyfire Mar 08 '15 at 15:52
  • @skyfire - you can see here more details and references. – Mauro ALLEGRANZA Mar 08 '15 at 16:14
  • As a followup on your answer, suppose we have ϕ appear at some point and don't discharge after the first appearance of implication introduction, and at some point later bring in another occurance of ϕ and apply implication introduction and this is the final line/conclusion, which occurrence of ϕ do we discharge? Does it not matter? I feel as tho 2.4.4 Does not adequately answer this because there was only one occurrence of ϕ with multiple applications of implication introduction. – skyfire Mar 08 '15 at 17:04