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I was given the following problem:

Find a derivation for$\{\varphi \Rightarrow (\psi \land \phi)\} \vdash \psi \Rightarrow (\varphi \Rightarrow \phi)$

The derivation is to be made using natural deduction. I came to the following derivation:

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However, I am unsure of the last step ($\Rightarrow I$). It is the case that $\psi$ was found to be true; namely, when we arrived to $\psi \land \phi$. So the conclusion seems correct from an intuitive point of view. But I think the formal application of the rule is incorrect. The rule for $\Rightarrow I$ states that

If $[\varphi] \ldots \phi $ is a derivation, then $\varphi \Rightarrow \phi$.

In our case, I should have gotten a derivation of the form $\psi \ldots (\varphi \Rightarrow \phi)$, but what I have is a derivation of the form $(\psi \land \phi) \ldots (\varphi \Rightarrow \phi)$, which is not identical to what is stated in the rule.

Of course, another rule is $\land E$, by virtue of which $\psi$ follows from $\psi \land \phi$, but I have never applied this rule explicitly for $\psi$ (only for $\phi$).

lafinur
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  • I think what you had is instead a derivation of the form $\varphi\ldots(\varphi\Rightarrow\phi)$, since you made an assumption on $\varphi$. If you want to deduce $\psi\Rightarrow (\varphi\Rightarrow \phi)$, you might have to make an assumption on $\psi$ first, based on the rule of implication introduction. – yanruijie Apr 29 '23 at 17:30
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    In general, the derivation $\phi \vdash \psi \to \phi$ is correct. – Mauro ALLEGRANZA Apr 29 '23 at 17:41
  • Thank you both for your comments.

    @MauroALLEGRANZA I can see why that is true. However, I am to justify each step of the derivation with a specific rule ($\land I, \to I, \to E$, etc.), and $\phi \vdash \psi \to \phi$ is a provable theorem but not a rule. If I am missing something, I apologize, I am fairly new to the subject.

     – lafinur Apr 29 '23 at 17:58
  • @yanruijie I was considering the same thing but was struggling with finding the correct point at which to assume $\psi$. Although I it is not very difficult to find it. What bothers me is the need to assume something that was already derived ($\psi$ was already found to follow from the assumption of $\varphi$) – lafinur Apr 29 '23 at 18:05
  • @lafinur I understand your point. However, to use implication introduction you must follow its syntax and assume $\psi$, and then assume $\varphi$. It looks redundant, but I believe it's necessary. – yanruijie Apr 29 '23 at 18:13
  • @yanruijie No, that's wrong. It is not necessary to explicitly state an assumption that is not actually needed. The case where the assumption doesn't actually occur is covered by the rule format $[\varphi] \ldots \phi$. – Natalie Clarius Apr 29 '23 at 23:17
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    The last step is a legal application of implication introduction where $\psi$ is not needed as an assumption. – Natalie Clarius Apr 29 '23 at 23:23
  • See this post – Mauro ALLEGRANZA May 02 '23 at 09:31
  • @MauroALLEGRANZA that was very useful thank you. – lafinur May 02 '23 at 18:22

1 Answers1

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However, I am unsure of the last step (⇒I). It is the case that $ψ$ was found to be true; namely, when we arrived to $ψ∧ϕ$. So, the conclusion seems correct from an intuitive point of view.

It does not matter whether you have derived the antecedent or not. When the consequent is derived, then the conditional can be derived too. (Anything will imply a truth.)

Thus, this is usually considered a valid application of the rule of conditional introduction.

$$\def\implies{\mathop{\Rightarrow}} \dfrac{\vdots\\q}{~p\implies q~}{\small(\implies\mathsf I)}$$

You may discharge any number assumptions of the antecedent occurring above the line, including none.

However some proof checkers do have stricter requirements, and so only accept a discharge of one and only one assumption,   Still, you may always add assumptions.

$$\dfrac{~\lower{2.5ex}{[p]^n}\quad{\vdots\\q}~}{p\implies q}{\small(\implies\mathsf I^n)}$$

But I think the formal application of the rule is incorrect. The rule for $⇒I$ states that

If $[\phi]\cdots\varphi$ is a derivation, then $\phi\implies\varphi$.

Unless your source gives alternatives, your system appears to insist on a strict implementation of the rule.   Your assignment marker may do so too.   If so, then you should add the assumption and discharge it.

Graham Kemp
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