Chapter Three, Tutorial One
Truth Tables
In this tutorial and the next we will show how to write a truth table for a sentence of SL. This will allow us to fully analyze the meaning of such a sentence. Later in this chapter, we will see how to apply truth tables to test, for instance, whether a sentence is logically true or an argument is valid.
To write truth tables for an arbitrary sentence of SL, you will have to know the tables which define the connectives. Below is one table for all five connectives; keep it in mind because it will be all important in the next few tutorials!
| P | Q | P&Q | PvQ | P>Q | P=Q | ~P |
| T | T | T | T | T | T | F |
| T | F | F | T | F | F | F |
| F | T | F | T | T | F | T |
| F | F | F | F | T | T | T |
Remember that P and Q are variables ranging over all SL sentences. Thus, the table applies to any sentence of our symbolic language.
Here's a very easy application of our table definition of the connectives. Suppose for definiteness that 'B' and 'C' continue to mean that Bob will attend law school and that Carola will attend law school and assume that both are TRUE. What can we say about '~B>C'?
First, notice that 'B' and 'C' are both assigned true: the 'T' underneath each indicates this.
What value should replace the question mark?
When you are comfortable with '~B>C' and why it is true, apply this same thinking and the truth table definitions of the connectives to the sentences below.
Continue to suppose that Bob and Carola will both attend law school. Click on all true sentences below:
T3.1: 2 of 3
You have just applied the truth table definitions of the connectives to several simple sentences given the truth values of their atomic components.
We will continue to suppose we know the truth values of atomic sentences, but consider a more complicated sentence.
(~Bv~C)>~B
Once a sentence gets the least bit complex, it's harder to evaluate its truth value. But we will do so by taking it one step at a time.
Start with the simplest sentences, the atomic B and C, given to be true. From these we determined that '~B' and '~C' are both false. So, '~Bv~C' is FALSE because of the definition of the 'v', right? (That's how you should have thought about it! A disjunction is false exactly when its disjuncts are both false.)
Bear with me! Almost finished. It follows from this -- '(~Bv~C)' false and '~B' false -- that '(~Bv~C)>~B' is true! Why? Because the text just highlighted says that both antecedent and consequent are FALSE. But our definition of the horseshoe says that in this case (row four, the 'F,F' possibility), the conditional is TRUE!
NOTE: The trick is to start with the shortest sentences (atomic ones) and work up in order of complexity just as you would build it. Then just apply the truth table definitions.