The Main Connective (Again)

Remember the main connective? Way back when we covered the rules of formation we introduced the notion of the main connective. Thinking about the main connective can be a very helpful way of dealing with longer compound sentences when doing derivations in natural deduction.

How does it help? Recognizing the main connective of a premise tells us what rules we can apply – it tells us how we can start off with our derivation. If the main connective of a premise is the conditional, for example, we know we can apply conditional elimination. Recognizing the main connective of the conclusion tells us what rules we will need to get to where we want to go. If the main connective is a conjunction, for example, we know we will probably need the conjunction introduction rule to get there.

Long compounds Say you are asked to prove the following argument: ((((P ∧ Q) ≡ (R ⊃ S)) ∨ ~~~(T ∧ Q)) ∨ ~(P ⊃ ~S)) ⊃ U (((A ∨ ~B) ∧ ~~C) ∨ ~~~D) ∧ ((P ∧ Q) ≡ (R ⊃ S)) ((E ≡ (F ∨ ~G)) ⊃ (H ∧ ~E)) ∨ (U ∨ ~~~(I ⊃ (J ∧ ~K))) It doesn’t look so nice and manageable.

Long compounds But by searching out that main connective we can make it all much more manageable. Take the first premise: ((((P ∧ Q) ≡ (R ⊃ S)) ∨ ~~~(T ∧ Q)) ∨ ~(P ⊃ ~S)) ⊃ U What is the main connective here? First let’s find the outermost set of brackets.

Long compounds ((((P ∧ Q) ≡ (R ⊃ S)) ∨ ~~~(T ∧ Q)) ∨ ~(P ⊃ ~S)) ⊃ U Now it’s not hard to spot that main connective. Because remember, the main connective is just the last connective that was applied in building the compound up from its components. And brackets tell us exactly what order the connectives were applied in.

Long compounds Our main connective is that final conditional. We’ve got that whole long compound ‘(((P ∧ Q) ≡ (R ⊃ S)) ∨ ~~~(T ∧ Q)) ∨ ~(P ⊃ ~S)’ as the antecedent, and that whole thing is related by the conditional to ‘U’ as the antecedent. The compound sentence is of the form ‘X ⊃ Y’ where X = (((P ∧ Q) ≡ (R ⊃ S)) ∨ ~~~(T ∧ Q)) ∨ ~(P ⊃ ~S), and Y = U

Long compounds So now we know premise one is of the form ‘X ⊃ Y’ we know we might be in a position to apply the conditional elimination rule. Because we know that the conditional rule applies to anything of that form. All we need is to have X on its own. So, let’s have a look at those premises again and see where we might get X from.

Long compounds Well, it looks like at least part of that initial X is going to be found in that second premise. So let’s have a look there. What’s the main connective of this second premise? Again, let’s look for the outermost brackets.

Long compounds (((A ∨ ~B) ∧ ~~C) ∨ ~~~D) ∧ ((P ∧ Q) ≡ (R ⊃ S)) We seem to have two separate pairs of outermost brackets here. And what lies outside of those brackets is our main connective!

Long compounds (((A ∨ ~B) ∧ ~~C) ∨ ~~~D) ∧ ((P ∧ Q) ≡ (R ⊃ S)) What we have here then is a sentence of the form ‘X ∧ Y’ where X = (((A ∨ ~B) ∧ ~~C) ∨ ~~~D) Y = ((P ∧ Q) ≡ (R ⊃ S)) And what does that tell us with regard to what rules we can apply?

Long compounds Well we have a conjunction here. And conjunction elimination tells us that when we have a conjunction like this, we can drop the connective and conclude either of the conjuncts. So from our second premise we can conclude ‘((P ∧ Q) ≡ (R ⊃ S))’ This is close to what we’re looking for, which is: (((P ∧ Q) ≡ (R ⊃ S)) ∨ ~~~(T ∧ Q)) ∨ ~(P ⊃ ~S)

Long compounds So let’s think about: (((P ∧ Q) ≡ (R ⊃ S)) ∨ ~~~(T ∧ Q)) ∨ ~(P ⊃ ~S) This sentence is still pretty long and confusing. So let’s give it the same treatment: where are the outermost brackets?

Long compounds (((P ∧ Q) ≡ (R ⊃ S)) ∨ ~~~(T ∧ Q)) ∨ ~(P ⊃ ~S) So our main connective is…

Long compounds (((P ∧ Q) ≡ (R ⊃ S)) ∨ ~~~(T ∧ Q)) ∨ ~(P ⊃ ~S) So we’ve got something of the form ‘X ∨ Y’ with X = (((P ∧ Q) ≡ (R ⊃ S)) ∨ ~~~(T ∧ Q)), and Y = ~(P ⊃ ~S)

Long compounds Remember where we are in the derivation. We’re trying to get from ‘((P ∧ Q) ≡ (R ⊃ S))’ (which we derived from our second premise), to ‘(((P ∧ Q) ≡ (R ⊃ S)) ∨ ~~~(T ∧ Q)) ∨ ~(P ⊃ ~S)’, which will allow us to use conditional elimination on our first premise. And we’ve now ascertained, through recognizing the main connective, that that second long compound is a disjunction.

Long compounds Well, if what we want is to get a disjunction, we’re probably going to try and use the disjunction introduction rule. Which says that for any X we may conclude that X ∨ Y. In the case of our long compound X = (((P ∧ Q) ≡ (R ⊃ S)) ∨ ~~~(T ∧ Q)), and Y = ~(P ⊃ ~S)

Long compounds So now we just need to get this shorter sentence: (((P ∧ Q) ≡ (R ⊃ S)) ∨ ~~~(T ∧ Q)) But look – the main connective of this sentence is a disjunction too! So if we want to get this sentence, again we just need disjunction introduction, applied to ‘((P ∧ Q) ≡ (R ⊃ S))’. And this is the sentence we already have – derived from our second premise!

Long compounds Ok, let’s take stock. Now we have worked out how to derive ‘U’ from our two premises. But what was to be proved was the longer sentence: ((E ≡ (F ∨ ~G)) ⊃ (H ∧ ~E)) ∨ (U ∨ ~~~(I ⊃ (J ∧ ~K))) We could go through all the steps just taken to try to work this out. But let’s try to just apply the thinking we’ve just gone through in detail a little more directly.

Long compounds What we have is ‘U’. And we can see that there is a ‘U’ there in our target sentence: ((E ≡ (F ∨ ~G)) ⊃ (H ∧ ~E)) ∨ (U ∨ ~~~(I ⊃ (J ∧ ~K))) Moreover, this ‘U’ is disjoined with ‘~~~(I ⊃ (J ∧ ~K))’ to make a disjunction. And then this whole disjunction is disjoined with ‘((E ≡ (F ∨ ~G)) ⊃ (H ∧ ~E))’ to make the full sentence. In other words, all we need to get from U to the target sentence is a couple of disjunctions!

Long compounds But this is easy – we can do that with disjunction introduction, which says that if we have X on a particular scope line, we may write X ∨ Y below on the same scope line. So we now have a route plotted out from the premises to the conclusion. We just need to put it all together into a derivation!

1. ((((P ∧ Q) ≡ (R ⊃ S)) ∨ ~~~(T ∧ Q)) ∨ ~(P ⊃ ~S)) ⊃ U. P. 2 1 ((((P ∧ Q) ≡ (R ⊃ S)) ∨ ~~~(T ∧ Q)) ∨ ~(P ⊃ ~S)) ⊃ U P 2 (((A ∨ ~B) ∧ ~~C) ∨ ~~~D) ∧ ((P ∧ Q) ≡ (R ⊃ S)) P 3 ((P ∧ Q) ≡ (R ⊃ S)) 2, ∧E 4 ((P ∧ Q) ≡ (R ⊃ S)) ∨ ~~~(T ∧ Q) 3, ∨I 5 ((((P ∧ Q) ≡ (R ⊃ S)) ∨ ~~~(T ∧ Q)) ∨ ~(P ⊃ ~S)) 4, ∨I 6 U 1, 5, ⊃E 7 U ∨ ~~~(I ⊃ (J ∧ ~K) 6, ∨I 8 ((E ≡ (F ∨ ~G)) ⊃ (H ∧ ~E)) ∨ (U ∨ ~~~(I ⊃ (J ∧ ~K))) 7, ∨I

The moral of the story Ok, that was a very long-winded example. But the moral of the story is relatively simple. When things get long and complex, just look for the main connective – and that should help a lot in terms of thinking about what to do next, and where you need to be heading towards. We can formalize these thoughts as follows.

The moral of the story The elimination rule for ‘∧’ applies to a sentence only when a ‘∧’ occurs as the sentence's main connective. The same thing goes for ‘∨’, ‘⊃’, and ‘≡’. The components used with the main connective are the components to which the elimination rule makes reference.

The moral of the story The elimination rule for ‘~’ applies only to a doubly negated sentence, ~~X; that is, only when ‘~’ is the sentence's main connective, and the ‘~’ is applied to a component, ~X, which itself has a ‘~’ as its main connective. The introduction rule for ‘∧’ licenses you to write as a conclusion a sentence, the main connective of which is ‘∧’. The same thing goes for ‘∨’, ‘⊃’, ‘≡’, and ‘~’.