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To show this proposition holds, I need to assume $\phi$ is a theorem, i.e. there is a derivation where the set of hypothesis is empty and its conclusion is $\phi$.

I need to prove $\psi \to \phi$ is a theorem, i.e. there is a derivation where the set of hypothesis is empty and its conclusion is $\psi \to \phi$.

Is my understanding correct ? How can I start filling the details of this proof ?

P.D.: The book I am using is "Logic and Structure" by Dirk Van Dalen.

F. Zer
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  • ISTR that if $\phi_1$, $\phi_2$,...$\phi_n=\phi$ is a valid proof of $\phi$, then you can show by induction that $\psi\to\phi_1$, $\psi\to\phi_2$, ..., $\psi\to\phi$ is a valid proof of $\psi\to\phi$. –  Oct 21 '20 at 22:05
  • You're talking about derivations in what system? – Taroccoesbrocco Oct 21 '20 at 22:07
  • This all depends on the inference rules you have. In fact, for some systems this is just not true... though I doubt van Dalen's system is one of those :P. Anyway, we need to know the inference rules that van Dalen uses to define $\vdash$ – Bram28 Oct 21 '20 at 22:09
  • I think that the set of hypotheses won't be empty, since $\vdash$$\phi$, and thus $\phi$ can be a hypothesis. Basically, assume $\psi$, get to $\phi$ and use $\rightarrow$I. A copy of Dirk Van Dalen's book, referenced in the original post, is here: https://cin.ufpe.br/~mlogica/livros/Logic%20and%20Structure%20-%20Van%20Dalen.pdf Rules start appearing on p. 30. – Doug Spoonwood Oct 21 '20 at 22:13
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    Thank you, everyone. @Taroccoesbrocco, the Natural Deduction system used by Van Dalen is Gentzen, and the rules of inference are Introduction and Elimination rules of conjunction, disjunction, implication and RAA, $\bot$. Is this what you are asking ? – F. Zer Oct 21 '20 at 22:15

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Yes, correct. You start with a derivation the end of which is $\phi$ and for which you assume it has no open assumptions, and argue that if this derivation exists, you can also find a derivation that ends in $\psi \to \phi$ with no further open assumption. When writing down the proof, you can use $\vdots$ or $\mathcal{D}$ on top of the conclusion to indicate the unknown start of the derivation.
For this particular proof (and many others of the form "If $\vdash A$ then $\vdash B$"), the argumentation that the other derivation exists will boil down showing that the derivation ending with $\phi$ can be continued, using the rules of ND, to reach the conclusion $\psi \to \phi$. Which step(s) that continuation will consist of should now be easy, if you correctly understood the business about rules with dischargeable assumptions.

Natalie Clarius
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  • Thank you so much, @lemontree ! My proof would be like this. Is it correct ? – F. Zer Oct 21 '20 at 23:51
  • I assume there exists a derivation of $\phi$ with no open assumptions. $$ \vdots\ \phi $$ Using the rule of $\to\mathbf{I}$, I can conclude $\psi \to \phi$ with no open assumptions. Therefore, $\vdash \psi \to \phi$. – F. Zer Oct 21 '20 at 23:52
  • Correct; well done! – Natalie Clarius Oct 21 '20 at 23:55
  • The point of this exercise (I think) is to have understood that $\to I$ (and other rules that allow to discharge assumptions) may be applied without discharging (all) occurrences of the assumptions, or (like here) without the assumption even present in the derivation at all. – Natalie Clarius Oct 21 '20 at 23:56
  • Thank you ! I think you raise something important, @lemontree. Could you clarify why you say that "$\to\mathbf{I}$ me be applied without discharging (all) occurrences of the assumptions" ? Is it something that could be deduced from rule definition ? Also, could you tell me a few rules that allow its application in a similar way (without discharging all occurrences of the assumptions) ?. – F. Zer Oct 22 '20 at 00:02
  • You already used this correctly in your proof. The rule ($\to I$) $\quad [\psi]^i \cdots \phi - \psi \to \phi\quad $ states that the antecedent $\psi$ to be introduced is an assumption from which $\phi$ was derived and which may be [discharged]${}^1$ when applying the rule. But in your proof, the assumption $\psi$ does not exist in the derivation of $\phi$, and we are allowed to $\to$-introduce it anyway: Dischargeable assumptions need not be actually present in the derivation. – Natalie Clarius Oct 22 '20 at 00:49
  • The other half of the issue is that existing occurrences of dischargeable assumptions need not actually be discharged: $\psi \cdots \phi - \psi \to \phi$, yielding $\psi \vdash \psi \to \phi$ (instead of $\vdash \psi \to \phi$) would also be a correct application of the $(\to I)$ rule. – Natalie Clarius Oct 22 '20 at 00:58
  • This holds for all rules that allow to discharge assumptions: You can either discharge all occurrences, leave all of them open, discharge some while keeping others open (talking about occurrences because it could be that the assumption was used multiple times in the derivation of the conclusion), or have no occurrence of the assumption at all. Other rules that allow to discharge assumptions are $\neg I$ (discharging the unnegated assumption), $\bot$ (discharging the negated assumption) and $\lor E$ (discharging the two disjunctions). – Natalie Clarius Oct 22 '20 at 01:02
  • That this is generally allowed is not obvious from the notation in the rule schemata themselves and probably was mentioned in the text when introducing the ND notation and explaining the discharging of assumptions. – Natalie Clarius Oct 22 '20 at 01:06
  • Thank you so much, @lemontree. I wasn't aware that this was allowed. May I ask, if I leave those assumptions undischarged, isn't the rule incorrectly applied ? Perhaps, I am missing the point. – F. Zer Oct 22 '20 at 01:19
  • Regarding your last comment, I found in the book on page 34: "With respect to the cancellation of hypotheses, we note that one does not necessarily cancel all occurrences of such a proposition ψ. This clearly is justified, as one feels that adding hypotheses does not make a proposition underivable (irrelevant information may always be added)." – F. Zer Oct 22 '20 at 01:19
  • I still do not see (although, continually use in proofs), why is the application of, for example, $\to I$ valid if I do not discharge (all) assumptions. For example, say I have an assumption 5 times, if I understand correctly, to use the inference in a correct way, I must discharge (at least) one of those assumption. But, what happens if there are none ? – F. Zer Oct 22 '20 at 01:22
  • Let us continue this discussion in chat. – F. Zer Oct 22 '20 at 13:12