# Truth Tables

Declarative word or number sentences are called **statements** if they are either true or false but cannot be both. Sometimes their truth values are easy to determine:

*There are 12 months in a year*is a true statement.- \(3+1=5\) is a false statement.
- \(n\) is an even number is not a statement (it is an open sentence) because the truth value depends on the value of \(n\).

Other times it might not be so obvious and a truth table may be needed. Truth tables show the truth values of one or more compound statements for each possible combination of truth values of the statements within them. In this video, we’ll see how they work with common logical operators, and then we’ll get into more complex conditional statements.

First up, negation, denoted by this symbol: ¬. Every time a statement \(P\) is negated, its truth value is reversed. The truth table shows the truth values resulting from different numbers of negations:

\(P\) | \(¬P\) | \(¬¬P\) | \(¬¬¬P\) | … |
---|---|---|---|---|

T | F | T | F | … |

F | T | F | T | … |

If the statement \(P\) is true, then the negation of \(P\) is false, the negation of the negation of \(P\) is true, the negation of the negation of the negation of \(P\) is false, and so on and so forth. If the statement \(P\) is false, then you will have the opposite truth values.

Suppose \(P\) is the statement, “*Snoopy is a dog*.” Using the truth table, since we can see this statement is true, we can also see that the statement, “*Snoopy is not not a dog*,” is also true.

\(P\) | \(¬P\) | \(¬¬P\) | \(¬¬¬P\) | … |
---|---|---|---|---|

T | F | T | F | … |

F | T | F | T | … |

If \(P\) is the statement, “*0.5 is an integer*,” which is false, the table shows that, “*0.5 is not an integer*,” is true.

\(P\) | \(¬P\) | \(¬¬P\) | \(¬¬¬P\) | … |
---|---|---|---|---|

T | F | T | F | … |

F | T | F | T | … |

Now for conjunction, joining two statements \(P\) and \(Q\) with the word “and”. The symbol for conjunction is an upside down V: ∧. A conjunction is true only when both statements are true, according to the table:

P | Q | P∧Q |
---|---|---|

T | T | T |

T | F | F |

F | T | F |

F | F | F |

If the statement \(P\) is true AND the statement \(Q\) is true, then \(P∧Q\) is true. If \(P\) is true and \(Q\) is false, then \(P∧Q\) is false. If \(P\) is false and \(Q\) is true, then \(P∧Q\) is false. And if \(P\) is false and \(Q\) is false, then \(P∧Q\) is false.

The statement, “*Snoopy is a dog and 0.5 is an integer*,” is false according to the table because \(P\) – *Snoopy is a dog* – is true, but \(Q\) – 0.5 is an *integer* – is false.

P | Q | P∧Q |
---|---|---|

T | T | T |

T | F | F |

F | T | F |

F | F | F |

Disjunction, two statements joined by “or” and represented by this symbol: \(∨\), has kind of an opposite effect. It is only false when both statements are false.

P | Q | P∨Q |
---|---|---|

T | T | T |

T | F | T |

F | T | T |

F | F | F |

If \(P\) is true and \(Q\) is true, then \(P∨Q\) is true. If \(P\) is true and \(Q\) is false, then \(P∨Q\) is true. If \(P\) is false and \(Q\) is true, then \(P∨Q\) is true. If \(P\) is false and \(Q\) is false, then \(P∨Q\) is also false.

The statement, “*Snoopy is a dog or 0.5 is an integer*,” is true according to the table because \(P\) – *Snoopy is a dog* – is true, and Q –* 0.5 is an integer* – is false.

Conditional statements, denoted by an arrow: →, can also be evaluated using truth tables. Consider the example, “*Natural numbers are integers*.” Of course, this can be written in “if, then” form: “*If a number is a natural number, then it is an integer*.”

Upon hearing this statement, you grab your calculator, ready to generate some random numbers. For each number generated, one of four things will happen:

- Your number will be a natural number that is also an integer.
- Your number will be a natural number that is not an integer.
- Your number will not be a natural number but will be an integer.
- Your number will not be a natural number and will not be an integer.

Scenario 1 could definitely be true. Scenario 2 could not. Scenarios 3 and 4, however, could go either way, but from the *statement only*, we don’t actually know anything about numbers that are not natural numbers. All we know is that if a number is a natural number, then it is an integer. So scenarios 3 and 4 don’t change the truth value of the conditional. Here’s what the truth table for a conditional looks like:

Scenario | P | Q | P→Q |
---|---|---|---|

1 | T | T | T |

2 | T | F | F |

3 | F | T | T |

4 | F | F | T |

Suppose I make this statement: “*If you make an A on your test, then I’ll give you 10 dollars*.” Is the statement, “If you make a B on your test, then I’ll give you 10 dollars,” true or false?

[pause]

P | Q | P→Q |
---|---|---|

T | T | T |

T | F | F |

F | T | T |

F | F | T |

The statement is true because we don’t know the outcome of bringing home a B. All we know is that it is true that an A will get us 10 bucks.

Think of it this way, if you got an A, then you should get $10. So when \(P\) is true (you make an A on your test) and \(Q\) is true (I give you $10), then \(P→Q\) is also true because the first thing caused the second thing to happen. I told you the truth. But what if you got an A on your test (\(P\) is true) and I didn’t give you $10 (\(Q\) is false)? Well, you’d probably be upset with me because I lied to you and told you a false statement (\(P→Q\) is false). These next two rows are where it gets a little tricky. What if you make a B on your test (\(P\) is false) and I give you $10 (\(Q\) is true). Is \(P→Q\) true? It may not seem like it, but it is true. I didn’t lie to you by giving you $10 even though you didn’t make an A because I didn’t tell you what I would do if you made a B. The same logic follows for the next statement. If you make a B on your test (\(P\) is false) and I don’t give you $10 (\(Q\) is false), then \(P→Q\) is still true. Again, I didn’t lie to you by not giving you $10 when you made a B. All I said was IF you made an A, then I would give you $10. There were no promises made or broken for you making a B.

The potential truth values of biconditionals can also be charted using truth tables. Biconditionals, usually containing “if and only if” and represented with a double arrow ↔, are true when both statements \(P\) and \(Q\) are either true or false. Definitions tend to work well as biconditionals. Here’s one: “A right angle measures 90 degrees.” This can also be written, “*An angle is a right angle if and only if it measures 90 degrees*.” Biconditionals go both ways, so the reverse also works: “*An angle measures 90 degrees if and only if it is a right angle*.”

Given an angle, there are four possible scenarios:

- An angle is a right angle and measures 90 degrees.
- An angle is a right angle and does not measure 90 degrees.
- An angle is not a right angle and measures 90 degrees.
- The angle is not a right angle and does not measure 90 degrees.

Clearly, scenarios 1 and 4 are true and 2 and 3 are false. Therefore, this is the truth table:

Scenario

Scenario | P | Q | P↔Q |
---|---|---|---|

1 | T | T | T |

2 | T | F | F |

3 | F | T | F |

4 | F | F | T |

Consider the statement “*Two lines are parallel if they are coplanar and do not intersect*.” Notice that the phrase “if and only if” has been simplified to if with two f’s. This is a common simplification made in mathematical phrases. Here, we’ll consider three statements:

- Two lines are parallel lines.
- Two lines are coplanar.
- Two lines do not intersect.

This statement is of the form \(P ↔ (Q∧R)\). Let’s make a truth table. Since there are three statements in this example and each one has its own true or false value, we know that our truth table will have \(2^3\) rows. In general, any truth table has \(2^n\) rows.

So the first three columns for \(P\), \(Q\), and \(R\) show the different possibilities for if \(P\), \(Q\), and \(R\) are true or false. Generally they’re set up the same way for any truth table that you have. Now we’re gonna look at the second part of our statement: \(Q∧R\).

For our first row, \(Q∧R\), we know that that’s true because \(Q\) and \(R\) are both true. Then we look at \(P\). So our statement \(P\) is true, and the statement \(Q∧R\) is true, so \(P ↔ (Q∧R)\) is also true.

Then for our second row we can do the same thing. Look at \(Q∧R\). \(R\) is false, so \(Q∧R\) is also false. Then we compare this to our statement \(P\). \(P\) is true and \(Q∧R\) is false, so \(P ↔ (Q∧R)\) is also false.

For our third row, we’re gonna look at \(Q∧R\). \(Q\) is false, so the statement \(Q∧R\) is false. Compared to our \(P\) statement which is true, that makes our biconditional statement, \(P ↔ (Q∧R)\) false because the \(Q∧R\) part is false.

For our fourth row, we’re gonna do the same thing, we look at \(Q∧R\), both \(Q\) and \(R\) are false, so \(Q∧R\) is also false. Compared to \(P\), which is true, that makes our biconditional statement, \(P ↔ (Q∧R)\) false because \(Q∧R\) is false, while \(P\) is true.

For our fifth row, we look at \(Q∧R\). Both \(Q\) and \(R\) are true, so \(Q∧R\) is true. Compared to our statement \(P\), which is false, \(P ↔ (Q∧R)\) is also false.

For our sixth row, \(Q∧R\) is false because \(R\) is false. Compared to \(P\), \(P\) is also false, which makes our biconditional statement, \(P ↔ (Q∧R)\) true because they have the same truth value. \(P\) has the same truth value as \(Q∧R\).

Looking at our seventh row, we’ll see the same thing. \(Q∧R\) is false because \(Q\) is false. But since \(P\) is also false, our biconditional statement becomes true. For our last row, we see that \(Q∧R \) is false because both \(Q\) and \(R\) have false truth values. And we compare this to our \(P\), which is also false, so \(P ↔ (Q∧R)\) is true because \(P\) is false and \(Q∧R\) is false, at the same time. Therefore, the whole thing is true.

P | Q | R | (Q∧R) | P ↔ (Q∧R) | Reasoning |
---|---|---|---|---|---|

T | T | T | T | T | “And” statement true, matches true p |

T | T | F | F | F | False R makes the “and” false but the P is true |

T | F | T | F | F | False Q makes the “and” false but P is true |

T | F | F | F | F | False Q and R make the “and” false but the P true |

F | T | T | T | F | P false, “and” true |

F | T | F | F | T | P and “and” both false |

F | F | T | F | T | P and “and” both false |

F | F | F | F | T | P and “and” both false |

Using the truth table, we can pinpoint whether the statement, “*Two lines are skew if they are coplanar and do not intersect*” is true or false. Skew lines, of course, do not intersect, but are not coplanar. Clearly a false statement, shown by the truth table.

P | Q | R | (Q∧R) | P ↔ (Q∧R) |
---|---|---|---|---|

T | T | T | T | T |

T | T | F | F | F |

T | F | T | F | F |

T | F | F | F | F |

F | T | T | T | F |

F | T | F | F | T |

F | F | T | F | T |

F | F | F | F | T |

Since we are considering skew lines, we will assume \(P\) is true. Skew lines are not coplanar, so \(Q\) is false. And skew lines do not intersect, so \(R\) is true. This gives us row 3, which shows that \((Q∧R)\) is false because \(Q\) is false, which means that \(P ↔ (Q∧R)\) is also false because \((Q∧R)\) is false, even though \(P\) is true.

I hope that this video helped you understand how truth tables work and how they can be used to determine truth values!

Thanks for watching, and happy studying!