# Why do the sequent calculus NOT left and NOT right rules work?

The rules I am considering are $\frac{\neg A, \ \Gamma \implies \Delta}{\Gamma \implies \Delta, \ A} (\neg L)$ and $\frac{\Gamma \implies \Delta, \ \neg A}{A, \ \Gamma \implies \Delta} (\neg R)$

I am trying to get my head around some of the sequent calculus rules, and while I think I understand most of them, I am struggling to apply any intuition to the negation rules shown above.

The intuition of looking at the left as a conjunction of literals and the right as a disjunction of literals seems to break down, and I am unclear how to explain these rules to myself.

What is a sensible way to view such rules and put some understanding on them?

• It becomes quite easy to understand once you combine two things: (1) the understanding that a sequent $A_1, ..., A_m \vdash B_1, ..., B_n$ can be seen as $(A_1 \wedge ... \wedge A_m) \rightarrow (B_1 \vee ... \vee B_n)$, together with (2) the translation of implication $A \rightarrow B$ to $\neg A \vee B$. May 20, 2013 at 11:00

You could start by considering simplified versions of the rules and build intuition by considering those cases. For instance,

$$\frac{\neg A, \ B \implies C}{B \implies C, \ A} (\neg L)$$

can be interpreted as stating that $(\neg A\wedge B)\Rightarrow C$ implies $B\Rightarrow (C\vee A)$.

So if it is the case that $\neg A$ and $B$ are true implies $C$ is true, then if only $B$ is true, either $C$ is true (independently of $\neg A$) or $A$ is true, for otherwise $\neg A$ would be true, which combined with $B$ being true would make $C$ true.

A similar game can be played with the other case.

Applying your intuition, the premise of $(\neg L)$ is: under the hypotheses $\Gamma$, and assuming $A$ is wrong, then one of the conclusions $\Delta$ holds. So if I only know $\Gamma$, what can I conclude?

• Possibly $\Delta$ holds.
• If $\Delta$ doesn't hold, it can only be that the assumption that $A$ is wrong is not met. In other words, $\neg (\neg A)$ is true, which means that $A$ is true.

All in all, from the premise of $(\neg L)$, we get that $\Gamma$ entails $\Delta$ or $A$. That's the conclusion of $(\neg L). Similarly, with$(\neg R)$: under the hypotheses$\Gamma$, either one of the conslusions$\Delta$holds, or$\neg A$holds. Now suppose that in addition to$\Gamma$we know that$A$is true. Then among the conclusions$\Delta, \neg A$, the item$\neg A$cannot be true, because you can't have both$A$and$\neg A$(excluded middle). So it must be one (or more) conclusions in$\Delta$that holds. All in all,$\Gamma$and$A$together entail$\Delta$, which is the conclusion of$(\neg R)$. You'll note that$(\neg L)$relies on the excluded middle. This rule does not hold in intuitionistic logic, where just because$\neg A$leads to a contradiction doesn't mean you can prove$A$.$(\neg R)\$ works in intuitionistic logic but might not in a fuzzy logic.