Conditional Introduction Rule

Question

In the derivation (the image below) the author shows that given the premise $\neg S \land \neg J$, the conclusion is $S \implies J$. All these deductive maneuver for concluding implications I find confusing. In this case you make an assumption that $\neg S \land \neg J$ (it implies that $S$ is negative), than you make an assumption that $S$ is positive (!). Then you ask: can $J$ be negative? You answer: no, there is a contradiction (which doesn't involve $J$), so $J$ must be positive. Then you conclude: if $S$ is positive, then $J$ is positive.

As I understand, if you assume $S$ and $\neg S$, you can prove that $X$ and $\neg X$ for any $X$: suppose we have $X$, look that it leads to a contradiction, we have $S$ and $\neg S$ at the same time, so we cannot have $X$, we must have $\neg X$. And by the same method we can show that $\neg X$also leads to a contradiction and we must have $X$.

Can you clarify this Conditional Introduction Rule (image below), which allows you to create assumptions that contradict to each other and from this conclude implications.

Answer 1

You have three rules in the "play" of the proof :

Conditional Introduction ($\rightarrow$-I; see in the top chart : $>$I) : if we have a derivation of $\psi$ from $\varphi$, then we can derive $(φ → ψ)$

Negation Elimination ($\lnot$-E) : if we have derived $\lnot \lnot \varphi$ , then we can derive $\varphi$ (also called : Double Negation)

Negation Introduction ($\lnot$-I) : if from $\varphi$ we have derived a contradiction, then we can derive $\lnot \varphi$.

Using Natural Deduction, we may state the above rules as follows [see Ian Chiswell & Wilfrid Hodges, Mathematical Logic (2007), page 17 and page 24] :

($\rightarrow$-I) $$\frac { \frac { [\varphi] } \psi } {(\varphi \rightarrow \psi)}$$

($\lnot$-E) $$\frac { \lnot \lnot \varphi } \varphi$$

($\lnot$-I) $$\frac { \frac { [\varphi] } \bot } { \lnot \varphi }$$

where $\bot$ stay for a contradiction.

This is the proof :

(1) $\lnot S \land \lnot J$ --- premise

(2) $S$ --- assumption $A_1$

(3) $\lnot J$ --- assumption $A_2$

(4) $\lnot S$ --- from (1) by $\land$-E

(5) $\lnot \lnot J$ --- from (3) and the contradiction (2)-(4) by $\lnot$-I, "discharging" assumption $A_2$

(6) $J$ --- from (5) by $\lnot$-E

(7) $S \rightarrow J$ --- from (2) and (6) by $\rightarrow$-I, "discharging" assumption $A_1$ --- conclusion.

As you can see, it is not Conditional Intro which "manages" contradictions; as per comment above, it licenses us to conclude from a derivation of $\psi$ under the assumtpion $\varphi$, to the conclusion $\varphi \rightarrow \psi$, which is nor more "dependent" of assumption $\varphi$.

Formally, the rule says that :

from the derivation : $\Gamma \cup \{ \varphi \} \vdash \psi$, we may obtain a new derivation : $\Gamma \vdash \varphi \rightarrow \psi$.

Here $\Gamma$ is a set (possibly empty) of assumptions which stay unchanged, while the assumption $\varphi$ of the initial derivation has been "discharged" in the new derivation.

The contradiction emerges from the "joint assertion" of the initial premise $\lnot S \land \lnot J$ and the two (temporary) assumptions : $S$ and $\lnot J$.

The contradiction tell us that the three cannot be asserted simultaneously; the rule for the management of contradictions ($\lnot$-I) give us the "liberty" to "blame" one of the three (we have to choose one) asserting its contradictory.

Thus, we decide to "blame" $\lnot J$, and "suppress" it from the set of assumptions, deriving its contradictory : $J$.

Having done this, we use Conditional Intro with the remaining (temporary) assumption : $S$, and we conclude with : $S \rightarrow J$.

The premise $\lnot S \land \lnot J$ "stay unchanged"; it is the only member of the set $\Gamma$.

Thus the last step is exactly : from the derivation $\Gamma \cup \{ S \} \vdash J$, obtain the new derivation : $\Gamma \vdash S \rightarrow J$, with $\Gamma = \{ \lnot S \land \lnot J \}$.