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The following is an example that my teacher came up with during one of our earlier lectures on natural deduction.

"Given the rules of inference that we have introduced so far, reflect on whether or not the following is the same implication-introduction rule that discharges assumptions:

$$ \frac{\varphi}{(\sigma \rightarrow \varphi)} \rightarrow I." $$

The rules of inference that we are working with states that, for implication-introduction to discharge assumptions one has to start with an assumption $[\sigma]$, then arrive at $\varphi$. The introduction of $\rightarrow$ then discharges the assumption (of course, given that we have only used the allowed rules of inference during the derivation).

This is clearly not the case here, as $\varphi$ has been presented with no derivation. However, arguing informally, this seems to make sense: if $\varphi$ is our assumption, then the implication will always hold, regardless of whether $\sigma$ is true or not. Perhaps this final "rule" is identical only if $\varphi$ is a premise?

Would love a detailed answer on this - thank you!

Azkar
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1 Answers1

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For any predicates $\sigma, \varphi$, when you derive that $\varphi$ is true, you may infer $\sigma\to\varphi$ holds.

Essentially, you are just immediately discharging an implicitly raised assumption (of $\sigma$).

Compare and contrast: $$\dfrac{\varphi}{\sigma\to\varphi}{\small({\to}\mathsf I)}\qquad\dfrac{[\sigma]^1\quad\varphi}{\sigma\to\varphi}{\small({\to}\mathsf I^1)}$$

The inference rule is often presented as follows, suggesting that the derivation of $\varphi$ must be relevant to the assumption of $\sigma$. However, this relevance is not required in Classical and Intuitionistic Logics. $\varphi$ may be derived independent of the assumption, including by being a premise.

$$\dfrac{\dfrac{[\sigma]^1\\~~\vdots}{\varphi}}{\sigma\to\varphi}{\small({\to}\mathsf I^1)}$$

Graham Kemp
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