Sometimes, it takes more than one
So far, we have looked at indirect truth tables that are resolved in one line. However, some indirect truth tables will need more than one line. Why? With some arguments, the initial assumption of invalidity will fail to reveal any definite information. In such cases, a second assumption will need to be made in order to determine validity. With such a secondary assumption, there will several different possibilities, or branches, to try. If, when trying such a branch, you fail to get a contradiction, then you know the argument is invalid. However, if you do get a contradiction, you do not know the argument is valid. You must go back and try another of the branches. Only if all branches generate a contradiction will the argument be shown invalid.

For example, consider the following argument:

P > Q
Q > S
P • S

The initial assumption of invalidity does not help us with this problem. For example, take the conclusion. Since our conclusion is a conjunction, assigning a false value to it does not reveal any definite information regarding P and S. We are unable to tell what the exact values of P and S will be— one of them, at least, must be false, but it could be either one or even both. Therefore, we have to consider each of these possibilities in turn. Each possibility represents a particular branch that might show the argument invalid.

The illustration above shows each of the three possible branches. There are three branches to consider because there are three different ways P • S might be false. All of the statements in the argument above offer three branches: there are three different ways each premise might be true. Given that, my preference is to work with the conclusion. However, if one of the premises had fewer branches than the conclusion, then that would be the best place to start. We must now try each branch. If a branch gives us a contradiction, that only shows that that branch fails to demonstrate invalidity. We must then try the next branch. Only if all three branches generate a contradiction will this argument be shown valid. On the other hand, if any of the branches fails to give us a contradiction, then that will show it is possible to assume the argument invalid. We can then stop without trying any remaining branches. Let us try branch one.

In branch one, we assume P is true and S is false. First, we apply those values to P and S in the rest of the problem. If P is true in the first premise, then Q must also be true. If Q is false, the first premise would be false. But if we make Q true, then in the second premise we have T > F. That gives us a false result. We have a contradiction and therefore must proceed to the second branch. (Remember: when we have branches a single contradiction does not guarantee validity. You have to get a contradiction on all the branches for the argument to be valid.)

The table above shows the contradiction. We must now look at the second branch. In that branch, we assume P false and S true. We must apply those values to our premises.

In the first premise, P is our antecedent. If the antecedent is false, what will the conditional statement be? It will be true. The value of Q does not matter. Both F > T and F > F are true. In our second premise, the consequent is true. That means the conditional statement is true, because both T > T and F > T are true. Again, the value of Q does not matter. So, in branch two, both premises are true without generating a contradiction. It is possible to assume the premises true and the conclusion false. The argument is therefore invalid. Since we know the argument can be assumed invalid, there is no need to try the third branch.

Another example:

~P > Q
Q > R
~R v D
P = D
D • P

Again, the initial assumption of invalidity reveals nothing about the individual variables. The best place to start is with the fourth premise, because there are only two ways for that statement to be true. That means we only have two branches to work with. In contrast, there are three ways the conclusion can be false. It would require three branches. The first three premises would also require 3 branches. So, P = D is the best statement to work with. Let us deal with the first branch.

The first branch immediately generates a contradiction. If P and D are both true, the conjunction of D • P is true. That makes the conclusion true. Our assumption of invalidity requires a false conclusion. That contradiction does not establish the argument is valid, however. It establishes only that the first branch does not work. We still have the second branch to try. Let us move on to it.

The second branch gives us a false conclusion, since both P and D are false. Moving to the first premise, if P is false, its negation is true. That makes the antecedent true. To get a true conditional statement, the consequent, Q, also must be true. Move to the second premise. Q is true. R cannot be false. That would give us T > F, which is false. So, R must be true. Moving to the third premise, R is negated. ~R is therefore false. But so is D. F v F is false. Yet the premise should be true. We again have a contradiction. Since we got a contradiction on all possible branches, that shows that this argument is valid.

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© 2003, David Arthur