Truth Tables for Arguments

1. Comparing Statements: We’ve looked at SINGLE propositions and assessed the truth values listed under their main operators to determine whether they were tautologous, self-contradictory, or contingent. In this section, we’re going to compare the truth tables of TWO propositions with EACH OTHER. To do this, we take the columns of truth vales under the main operators for EACH of the two propositions, and COMPARE them.

Sometimes, when we compare the truth tables for two different propositions, the truth values under the main operator will be exactly the same (when this happens, the two propositions are said to be “logically equivalent”). Sometimes they will be exactly opposite (when this happens, the two propositions are said to be “contradictory”). Sometimes, there will be one or more lines where both propositions come out true (in this case, the propositions are “consistent”). Sometimes, there will NOT be any lines where both propositions are true (in this case, the propositions are “inconsistent”).

Columns Under the Main Operators Relation Between SAME Truth Values on EVERY Line Logically Equivalent DIFFERENT Truth Values on EVERY Line Contradictory At Least One Line Where Both Are True Consistent No Line Where Both Are True Inconsistent

Let’s look at an example. Remember when we said that we can translate “unless” statements in one of THREE different ways—it doesn’t matter with because they all mean the same thing? For instance, we said that the statement…

“You will Lose the Lottery unless you Have a ticket.” (“L unless H”)

…can be translated in ANY one of the following three ways:

(1) “If you don’t have a ticket, then you will lose the lotto.” Translation: ¬H  L

(2) “If you will win the lotto, then you have a ticket.” Translation: ¬L  H

(3) “Either you will lose the lottery, or you have a ticket.” Translation: L  H

Let’s do truth tables for all three of these statements. Start with the first two. I’ll use a truth table chart for this one, like this:

L H ¬H  L ¬L  H T T T T T T T F F T T F F T T F F T F F F F F F

I’ve filled in the truth values of ‘L’ and ‘H’ under each letter. Let’s do the negations now:

L H ¬H  L ¬L  H T T FT T FT T T F TF T FT F F T FT F TF T F F TF F TF F

Finally, let’s find the final truth values of each statement under the main operator ():

L H ¬H  L ¬L  H T T FT T T FT T T T F TF T T FT T F F T FT T F TF T T F F TF F F TF F F

The statements are false in the fourth line because a conditional (“if … then”) claim is always true EXCEPT when the antecedent is true and the consequent is false. But notice something. Looking at the red truth values under each statement, these two statements have exactly the same truth table. Since their truth values are the same on all four lines, we say that they are logically equivalent. Now, recall the truth table for the ‘’ operator:

L H L  H T T T T F T F T T F F F

IT TOO has the same truth table! This means that ALL THREE of the following statements are logically equivalent—or in other words, they all mean exactly the same thing:

¬H  L  ¬L  H  L  H

So, we’ve learned that it IS permissible to translate “L unless H” in all three ways. 2 Note on ‘only if’: Recall that when we translated only if statements such as, “You will win the lottery only if you have a ticket”, we said that both ‘W  H’ as well as ‘¬H  ¬W’ were acceptable. Draw up truth tables for each to see that these are logically equivalent.

W  H  ¬H  ¬W

Let’s do another example. Imagine that an economist claims: “The balance of payments will decrease if and only if interest rates remain steady; however, it is not the case that either interest rates will not remain steady or the balance of payments will decrease.”

There are two claims here. (The second claim begins with the word “however”) Let “D”=”The balance of payments will decrease, and let “S”=”interest rates remain steady”. Then, we get the following two statements:

D  S ¬(¬S  D)

To compare the two propositions, we will need to do TWO truth tables—one for each statement. Since there are two letters, our truth table will have four lines. Like this:

D  S ¬(¬S  D) T T T T T F F T F T T F F F F F

Next, we can completely finish the first statement on the left (it’s just the truth table for “”). For the one on the right, let’s start by solving for the “¬” in front of the “S”:

D  S ¬(¬S  D) T T T FT T T F F TF T F F T FT F F T F TF F

The negations are in green. Next, let’s solve the disjunction by comparing the green letters with the black letters underneath the “D”. Remember that a disjunction is only false when BOTH disjuncts are false. So, we get the following:

D  S ¬(¬S  D) T T T FT T T T F F TF T T F F T FT F F F T F TF T F 3 Finally, we should solve for the final negation, OUTSIDE of the parenthesis. We do this by taking the blue letters and writing down the OPPOSITE of what is written in blue:

D  S ¬(¬S  D) T T T F FT T T T F F F TF T T F F T T FT F F F T F F TF T F

Now that we have the truth tables completed for both propositions, we can compare them. To do this, we simply compare the columns in red with one another. Like this:

D  S ¬(¬S  D) T F F F F T T F

Are the two statements logically equivalent? No. To be logically equivalent, the truth values in red would need to be IDENTICAL. Are they contradictory? No. To be contradictory, the truth values in red would need to be EXACTLY OPPOSITE. Though lines 1, 3, and 4, are exactly opposite, on line 2 they are BOTH FALSE (F). Are they consistent? No. To be consistent, there would need to be at least one line where BOTH statements are true. But there are NO lines with two T’s.

Answer: The economist’s two statements are inconsistent. That is, there is NO circumstance in which both propositions could be true; i.e., there is NO line with two T’s.

2. Truth Tables for 2-Letter Arguments: We have learned how to make truth tables for propositions. We have even learned how to construct truth tables to compare two propositions. In this section, we will continue to construct truth tables for multiple propositions—but now we’ll be constructing them in order to determine whether an argument is valid or invalid. Let’s start with an argument which has ONLY TWO propositions as components (one premise and a conclusion).

Example #1: For instance, imagine that I told you: “If you don’t study, then you will not get good grades. Therefore, if you do study, then you will get good grades.” Let “S”=”You study” and “G”=”You will get good grades”. In that case, my argument can be written as the following:

Premise: 1. ¬S  ¬G Conclusion: 2.  S  G

4 Is this argument valid or invalid? In order to answer that question, we will need to draw up truth tables for BOTH the premise AND the conclusion. Before we start, we first write up a four line truth table, with truth values for S and G, like this:

S G ¬S  ¬G S  G T T ? ? T F ? ? F T ? ? F F ? ?

The shading represents the divide between the premise and the conclusion. Recall our definition of the term “validity”: We said that a valid argument is one for which it is impossible for the premises to be true and the conclusion false.

We then used “the counter-example method” to determine whether or not an argument is invalid. If we could come up with an argument with the same FORM, where the premises were obviously true and the conclusion was obviously false, then we concluded that the argument was INVALID.

That is basically what we’ll be doing here, but with truth tables. We will write up truth tables for the premises and the conclusion, and if there is any line where the premises are ALL true, and the conclusion is false, then the argument is invalid. Otherwise (if there is no such line where this occurs), the argument is valid.

Let’s do the truth table for the argument above now. First, we simply fill in the truth values under S and G, like this:

S G ¬S  ¬G S  G T T T T T T T F T F T F F T F T F T F F F F F F

Next, let’s get rid of the negations in the premise, like this:

S G ¬S  ¬G S  G T T FT F T T T T F FT T F T F F T TF F T F T F F TF T F F F

5 Finally, we can solve for the conditionals (“”). Remember that a conditional is ALWAYS true unless it has a true antecedent and a false consequent.

S G ¬S  ¬G S  G T T FT T F T T T T T F FT T T F T F F F T TF F F T F T T F F TF T T F F T F

Now, in order to determine whether this argument is valid or invalid, we simply look for a line where the premise(s) are true and the conclusion is false. Look at each line. Is there any line where the red letter on the left is a “T“ and the letter on the right is an “F”? There sure is! The second line does this:

S G ¬S  ¬G S  G T T FT T FT T T T T F FT T TF T F F F T TF F FT F T T F F TF T TF F T F

Because there IS a line where the premise is true and the conclusion is false (circled in red above), this argument is INVALID. In other words, even if it is true that those who don’t study will get bad grades, this does NOT entail that those who DO study WILL get GOOD grades. The inference is invalid.

3. Truth Tables for 3-Letter Arguments: Let’s try a more complicated argument with MULTIPLE premises and THREE different statement letters:

Example #2: An employer says, “If racial quotas are adopted for promoting employees, then qualified employees will be passed over; but if racial quotas are not adopted, then prior discrimination will go unaddressed. Either racial quotas will or will not be adopted for promoting employees. Therefore, either qualified employees will be passed over or prior discrimination will go unaddressed.”

Let “R”=”Racial quotas are adopted”, “Q”=”Qualified employees are passed over”, and “P”=”Prior discrimination goes unaddressed”. In that case, the argument can be written as the following:

1. R  Q 2. ¬R  P 3. R  ¬R 4.  Q  P

6 The truth table to be filled in will look like the following:

R P Q R  Q ¬R  P R  ¬R Q  P T T T ? ? ? ? T T F ? ? ? ? T F T ? ? ? ? T F F ? ? ? ? F T T ? ? ? ? F T F ? ? ? ? F F T ? ? ? ? F F F ? ? ? ?

Let’s start by filling in all of the truth values for R, P, and Q, like this:

R P Q R  Q ¬R  P R  ¬R Q  P T T T T T T T T T T T T T F T F T T T T F T T F T T T T F T T T F T F F T F T F T T F F F T T F T F T F F T T F T F F F F T F F F T F F T F T F F F F T F F F F F F F F F F F F

Next, let’s get rid of the negations in the second and third premise, like this:

R P Q R  Q ¬R  P R  ¬R Q  P T T T T T FT T T FT T T T T F T F FT T T FT F T T F T T T FT F T FT T F T F F T F FT F T FT F F F T T F T TF T F TF T T F T F F F TF T F TF F T F F T F T TF F F TF T F F F F F F TF F F TF F F

Next, we can solve for all of the premises and the conclusion. The first two premises are conditionals (which are ONLY false when they have a true antecedent and a false consequent), while the third premise and the conclusion are disjunctions (which are ONLY false when BOTH disjuncts are false). Here is the result:

R P Q R  Q ¬R  P R  ¬R Q  P T T T T T T FT T T T T FT T T T T T F T F F FT T T T T FT F T T T F T T T T FT T F T T FT T T F T F F T F F FT T F T T FT F F F F T T F T T TF T T F T TF T T T F T F F T F TF T T F T TF F T T F F T F T T TF F F F T TF T T F F F F F T F TF F F F T TF F F F

Now we look for lines where the conclusion is false and the premises are true. There are only 2 lines where the conclusion is false, so we should only be concerned with those:

The two yellow highlighted lines are the ONLY lines where the conclusion is false. We should ignore all of the others. Now ask, are either of those lines ones where ALL of the premises are true?

Nope… On the first yellow line, premise 1 (R  Q) is false. On the second yellow line, the second premise (¬R  P) is false. So, since there ARE NOT any lines where all of the premises are true and the conclusion is false, the argument is valid.