Truth Tables and Validity

With a truth table, we can determine whether or not an ** argument**
is

Suppose we wish to determine whether the argument '(P → Q) and P, therefore Q' is valid. The first thing to do is to construct a truth table that will allow us to test the argument for validity. In the examples on this page. I am collapsing several steps, so if you are a bit puzzled, review the basic pages on truth tables and equivalence tests.

**(STEP 1)** Construct the basic truth table for the
argument, including the necessary guide columns.
Since the argument contains only 2 statement letters, P and Q, we need a 4 row
truth table.

1 | 2 | 3 | 4 | 5 | 6 | 7 |

P | Q | (P | → | Q) | P |
∴Q |

T | T | |||||

T | ⊥ | |||||

⊥ | T | |||||

⊥ | ⊥ |

Ultimately we will be examining each row to see whether
there is any row in which the value for Columns 4 and 6 is True while the
value for Column 7 is false. Notice that I have used a thicker line between columns 5 and 6
and between columns 7 and 8 to indicate the
break between the between the premises and the conclusion. The turnstile
(**∴**) before
the Q indicate that Q is the conclusion of the argument.

**(STEP ) **We now want to construct a truth table for
our particular sentence. The first step is to assign truth values to each
sentence letter in our premises--values that come directly from the guide
columns. We do the same for the conclusion.

1 | 2 | 3 | 4 | 5 | 6 | 7 |

P | Q | (P | → | Q) | P |
∴Q |

T | T | T | T | T | T | |

T | ⊥ | T | ⊥ | T | ⊥ | |

⊥ | T | ⊥ | T | ⊥ | T | |

⊥ | ⊥ | ⊥ | ⊥ | ⊥ | ⊥ |

**(STEP 4)** All that is left to do now is to fill in
the remaining values for the compound statements and then make a determination
as to the validity of the argument.

1 | 2 | 3 | 4 | 5 | 6 | 7 |

P | Q | (P | → | Q) | P |
∴Q |

T | T | T | T | T | T | T |

T | ⊥ | T | ⊥ | ⊥ | T | ⊥ |

⊥ | T | ⊥ | T | T | ⊥ | T |

⊥ | ⊥ | ⊥ | T | ⊥ | ⊥ | ⊥ |

Since there is no row win which all of the premises are true and the conclusion false, this argument is valid.

Compare the previous argument to a similar looking argument, and one that many people who have not had the benefit of a class in formal logic mistakenly believe to be valid:

(P → Q) and not P, therefore not Q

**(STEP 1)** Construct the basic truth table for the
argument, including the necessary guide columns.
Since the argument contains only 2 statement letters, P and Q, we need a 4 row
truth table.

1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |

P | Q | (P | → | Q) | ~ |
P |
∴~ |
Q |

T | T | |||||||

T | ⊥ | |||||||

⊥ | T | |||||||

⊥ | ⊥ |

Ultimately we will be examining each row to see whether there is any row in which the value for Columns 4 and 6 is True while the value for Column 8 is false. Notice that I have used a thicker line between columns 5 and 6 and between columns 7 and 8 to indicate the break between the between the premises and the conclusion. Since the conclusion of the argument is the negation of a statement, Column 8 is the column for the dominant operator in the conclusion.

* (Step 2)* Fill in the values for the
simple statements in the premises and the conclusion:

1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |

P | Q | (P | → | Q) | ~ |
P |
∴~ |
Q |

T | T | T | T | T | T | |||

T | ⊥ | T | ⊥ | T | ⊥ | |||

⊥ | T | ⊥ | T | ⊥ | T | |||

⊥ | ⊥ | ⊥ | ⊥ | ⊥ | ⊥ |

** (Step 3)** Fill in the values for the
compound statements and determine whether the argument is valid.

1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |

P | Q | (P | → | Q) | ~ |
P |
∴~ |
Q |

T | T | T | T | T | ⊥ | T | ⊥ | T |

T | ⊥ | T | ⊥ | ⊥ | ⊥ | T | T | ⊥ |

⊥ |
T |
⊥ |
T |
T |
T |
⊥ |
⊥ |
T |

⊥ |
⊥ |
⊥ |
T |
⊥ |
T |
⊥ |
T |
⊥ |

In Row 3 of this truth table, all of the premises are true and the conclusion is false, so the argument is NOT VALID.