We’ve seen how to calculate the truth value of a compound proposition from the truth values of its components. Sometimes we can do the opposite, working backward to calculate the truth value of each sentence letter from the truth value of the compound proposition as a whole.
If Albert showed up at the party, then unless Betty and Clara were both there, Darla left early.
Given that the above sentence is false, can you determine who was at the party and whether Darla left early? To find out, let’s begin by symbolizing the sentence in propositional logic, using the following sentence letters:
The italicized proposition above can be symbolized like this:
(A ⊃ ((B • C) ∨ D))The proposition as a whole is false, which means that its main connective, the conditional (symbolized ‘⊃’), is false. There is only one way for a conditional to be false, namely when its antecedent is true and its consequent is false. So, A is true and ((B • C) ∨ D) is false. Moreover, there is only one way for the disjunction ((B • C) ∨ D) to be false, namely if both disjuncts are false. So, (B • C) is false and D is false.
To make this calculation process easier, we can write the truth values below each of the letters and connectives, as we did previously when calculating the truth value of a compound proposition.
Step 1. The truth values of the conditional is given, so we write it beneath the ‘⊃” symbol:| ( | A | ⊃ | (( | B | • | C | ) | ∨ | D | )) |
| 0 |
| ( | A | ⊃ | (( | B | • | C | ) | ∨ | D | )) |
| 1 | 0 | 0 |
| ( | A | ⊃ | (( | B | • | C | ) | ∨ | D | )) |
| 1 | 0 | 0 | 0 | 0 |
So far, we have determined that Albert did show up, and that Darla didn’t leave early.
We cannot determine who else was at the party, because there are several ways for (B • C) to be false: perhaps Betty and Clara both missed the party, or perhaps only Betty missed it, or perhaps only Clara did. But we do know that they weren’t both there. This means that there are exactly three possible combinations of values that B and C could have, given that the proposition as a whole is false. (If we constructed a truth table for the above proposition, we would find that the proposition is false on exactly 3 rows.) We can represent those three possibilities as follows:
| ( | A | ⊃ | (( | B | • | C | ) | ∨ | D | )) |
| 1 | 0 | 0 | 0 | 0 | 0 | 0 | ||||
| 1 | 0 | 0 | 0 | 1 | 0 | 0 | ||||
| 1 | 0 | 1 | 0 | 0 | 0 | 0 |
As illustrated in the above example, it isn’t always possible to calculate the values of all the sentence letters in a compound proposition. Nevertheless, even when we can’t determine the values of all the letters, we can at least narrow down the possibilities. This will prove quite useful, as we’ll see later in this chapter.