Section 01: Basic rules for SL
In designing a proof system, we could just start with disjunctive syllogism and modus ponens. Whenever we discovered a valid argument which could not be proven with rules we already had, we could introduce new rules. Proceeding in this way, we would have an unsystematic grab bag of rules. We might accidently add some strange rules, and we would surely end up with more rules than we need.
Instead, we will develop what is called a natural deduction system. In a natural deduction system, there will be two rules for each logical operator: an introduction rule that allows us to prove a sentence that has it as the main logical operator and an elimination rule that allows us to prove something given a sentence that has it as the main logical operator.
In addition to the rules for each logical operator, we will also have a reiteration rule. If you already have shown something in the course of a proof, the reiteration rule allows you to repeat it on a new line. For instance:
When we add a line to a proof, we write the rule that justifies that line. We also write the numbers of the lines to which the rule was applied. The reiteration rule above is justified by one line, the line that you are reiterating. So the ‘R 1’ on line 2 of the proof means that the line is justified by the reiteration rule (R) applied to line 1.
Obviously, the reiteration rule will not allow us to show anything new . For that, we will need more rules. The remainder of this section will give introduction and elimination rules for all of the sentential connectives. This will give us a complete proof system for SL. Later in the chapter, we introduce rules for quantifiers and identity.
Conjunction
Think for a moment: What would you need to show in order to prove \(E\) & \(F\)?
Of course, you could show \(E\) & \(F\) by proving \(E\) and separately proving \(F\). This holds even if the two conjuncts are not atomic sentences. If you can prove [(\(A\)∨J) → \(V\) ] and [(\(V\) → \(L\)) ↔ (\(F\) ∨ \(N\))], then you have effectively proven
[(\(A\)∨J) → \(V\) ]&[(\(V\) → \(L\)) ↔ (\(F\) ∨ \(N\))].
So this will be our conjunction introduction rule, which we abbreviate &I:
A line of proof must be justified by some rule, and here we have ‘&I \(m\), \(n\).’ This means: Conjunction introduction applied to line \(m\) and line \(n\). These are variables, not real line numbers; \(m\) is some line and \(n\) is some other line. In an actual proof, the lines are numbered 1,2,3,... and rules must be applied to specific line numbers. When we define the rule, however, we use variables to underscore the point that the rule may be applied to any two lines that are already in the proof. If you have \(K\) on line 8 and \(L\) on line 15, you can prove (\(K\) & \(L\)) at some later point in the proof with the justification ‘&I 8, 15.’
Now, consider the elimination rule for conjunction. What are you entitled to conclude from a sentence like \(E\) & \(F\)? Surely, you are entitled to conclude \(E\); if \(E\) & \(F\) were true, then \(E\) would be true. Similarly, you are entitled to conclude \(F\). This will be our conjunction elimination rule, which we abbreviate &E:
When you have a conjunction on some line of a proof, you can use &E to derive either of the conjuncts. The &E rule requires only one sentence, so we write one line number as the justification for applying it.
Even with just these two rules, we can provide some proofs. Consider this argument.
The main logical operator in both the premise and conclusion is conjunction. Since conjunction is symmetric, the argument is obviously valid. In order to provide a proof, we begin by writing down the premise. After the premises, we draw a horizontal line— everything below this line must be justified by a rule of proof. So the beginning of the proof looks like this:
From the premise, we can get each of the conjuncts by &E. The proof now looks like this:
The rule &I requires that we have each of the conjuncts available somewhere in the proof. They can be separated from one another, and they can appear in any order. So by applying the &I rule to lines 3 and 2, we arrive at the desired conclusion. The finished proof looks like this:
This proof is trivial, but it shows how we can use rules of proof together to demonstrate the validity of an argument form. Also: Using a truth table to show that this argument is valid would have required a staggering 256 lines, since there are eight sentence letters in the argument.
Disjunction
If \(M\) were true, then \(M\)∨\(N\) would also be true. So the disjunction introduction rule (∨\(I\)) allows us to derive a disjunction if we have one of the two disjuncts:
Notice that \(\mathcal{B}\) can be any sentence whatsoever. So the following is a legitimate proof:
It may seem odd that just by knowing \(M\) we can derive a conclusion that includes sentences like \(A\), \(B\), and the rest— sentences that have nothing to do with \(M\). Yet the conclusion follows immediately by ∨I. This is as it should be: The truth conditions for the disjunction mean that, if \(\mathcal{A}\) is true, then \(\mathcal{A}\)∨\(\mathcal{B}\) is true regardless of what \(\mathcal{B}\) is. So the conclusion could not be false if the premise were true; the argument is valid.
Now consider the disjunction elimination rule. What can you conclude from \(M\) ∨ \(N\)? You cannot conclude \(M\). It might be \(M\)’s truth that makes \(M\) ∨ \(N\) true, as in the example above, but it might not. From \(M\)∨\(N\) alone, you cannot conclude anything about either \(M\) or \(N\) specifically. If you also knew that \(N\) was false, however, then you would be able to conclude \(M\).
This is just disjunctive syllogism, it will be the disjunction elimination rule (∨\(E\)).
Conditional
Consider this argument:
The argument is certainly a valid one. What should the conditional introduction rule be, such that we can draw this conclusion?
We begin the proof by writing down the premise of the argument and drawing a horizontal line, like this:
If we had¬\(R\) as a further premise, we could derive \(F\) by the ∨E rule. We do not have ¬\(R\) as a premise of this argument, nor can we derive it directly from the premise we do have— so we cannot simply prove \(F\). What we will do instead is start a subproof , a proof within the main proof. When we start a subproof, we draw another vertical line to indicate that we are no longer in the main proof. Then we write in an assumption for the subproof. This can be anything we want. Here, it will be helpful to assume ¬\(R\). Our proof now looks like this:
It is important to notice that we are not claiming to have proven ¬\(R\). We do not need to write in any justification for the assumption line of a subproof. You can think of the subproof as posing the question: What could we show if ¬\(R\) were true? For one thing, we can derive \(F\). So we do:
This has shown that if we had ¬\(R\) as a premise, then we could prove \(F\). In effect, we have proven ¬\(R\) → \(F\). So the conditional introduction rule (→I) will allow us to close the subproof and derive ¬\(R\) → \(F\) in the main proof. Our final proof looks like this:
Notice that the justification for applying the →I rule is the entire subproof. Usually that will be more than just two lines.
It may seem as if the ability to assume anything at all in a subproof would lead to chaos: Does it allow you to prove any conclusion from any premises? The answer is no, it does not. Consider this proof:
It may seem as if this is a proof that you can derive any conclusions \(\mathcal{B}\) from any premise \(\mathcal{A}\). When the vertical line for the subproof ends, the subproof is closed . In order to complete a proof, you must close all of the subproofs. And you cannot close the subproof and use the R rule again on line 4 to derive \(\mathcal{B}\) in the main proof. Once you close a subproof, you cannot refer back to individual lines inside it.
Closing a subproof is called discharging the assumptions of that subproof. So we can put the point this way: You cannot complete a proof until you have discharged all of the assumptions besides the original premises of the argument.
Of course, it is legitimate to do this:
This should not seem so strange, though. Since \(\mathcal{B}\)→\(\mathcal{B}\) is a tautology, no particular premises should be required to validly derive it. (Indeed, as we will see, a tautology follows from any premises.)
Put in a general form, the →I rule looks like this:
When we introduce a subproof, we typically write what we want to derive in the column. This is just so that we do not forget why we started the subproof if it goes on for five or ten lines. There is no ‘want’ rule. It is a note to ourselves and not formally part of the proof.
Although it is always permissible to open a subproof with any assumption you please, there is some strategy involved in picking a useful assumption. Starting a subproof with an arbitrary, wacky assumption would just waste lines of the proof. In order to derive a conditional by the →I, for instance, you must assume the antecedent of the conditional in a subproof.
The →I rule also requires that the consequent of the conditional be the last line of the subproof. It is always permissible to close a subproof and discharge its assumptions, but it will not be helpful to do so until you get what you want.
Now consider the conditional elimination rule. Nothing follows from \(M\) → \(N\) alone, but if we have both \(M\) → \(N\) and \(M\), then we can conclude \(N\). This rule, modus ponens, will be the conditional elimination rule (→E).
Now that we have rules for the conditional, consider this argument:
We begin the proof by writing the two premises as assumptions. Since the main logical operator in the conclusion is a conditional, we can expect to use the →I rule. For that, we need a subproof— so we write in the antecedent of the conditional as assumption of a subproof:
We made \(P\) available by assuming it in a subproof, allowing us to use →E on the first premise. This gives us \(Q\), which allows us to use →E on the second premise. Having derived \(R\), we close the subproof. By assuming \(P\) we were able to prove \(R\), so we apply the →I rule and finish the proof.
Biconditional
The rules for the biconditional will be like double-barreled versions of the rules for the conditional.
In order to derive \(W\) ↔ \(X\), for instance, you must be able to prove \(X\) by assuming \(W\) and prove \(W\) by assuming \(X\). The biconditional introduction rule (↔I) requires two subproofs. The subproofs can come in any order, and the second subproof does not need to come immediately after the first— but schematically, the rule works like this:
The biconditional elimination rule (↔E) lets you do a bit more than the conditional rule. If you have the left-hand subsentence of the biconditional, you can derive the right-hand subsentence. If you have the right-hand subsentence, you can derive the left-hand subsentence. This is the rule:
Negation
Here is a simple mathematical argument in English:
Assume there is some greatest natural number. Call it \(A\).
That number plus one is also a natural number.
Obviously, \(A\) + 1 > \(A\).
So there is a natural number greater than \(A\).
This is impossible, since \(A\) is assumed to be the greatest natural number.
.˙. There is no greatest natural number.
This argument form is traditionally called a reductio . Its full Latin name is reductio ad absurdum , which means ‘reduction to absurdity.’ In a reductio, we assume something for the sake of argument— for example, that there is a greatest natural number. Then we show that the assumption leads to two contradictory sentences— for example, that \(A\) is the greatest natural number and that it is not. In this way, we show that the original assumption must have been false.
The basic rules for negation will allow for arguments like this. If we assume something and show that it leads to contradictory sentences, then we have proven the negation of the assumption. This is the negation introduction (¬I) rule:
For the rule to apply, the last two lines of the subproof must be an explicit contradiction: some sentence followed on the next line by its negation. We write ‘for reductio’ as a note to ourselves, a reminder of why we started the subproof. It is not formally part of the proof, and you can leave it out if you find it distracting.
To see how the rule works, suppose we want to prove the law of non-contradiction: ¬(\(G\)&¬\(G\)). We can prove this without any premises by immediately starting a subproof. We want to apply ¬I to the subproof, so we assume (\(G\)&¬\(G\)). We then get an explicit contradiction by &E. The proof looks like this:
The ¬E rule will work in much the same way. If we assume ¬\(\mathcal{A}\) and show that it leads to a contradiction, we have effectively proven \(\mathcal{A}\). So the rule looks like this: