# Level 4.1: Proofs of Common Forms

As we saw in Level 2: Logical Properties & Relations course, truth-tables can be used to test individual formulae for necessary truth, necessary falsehood and contingency, and to test pairs of formulae for equivalence, consistency and contradiction, and entailment. And in the Proof by Analogy course, we were introduced to the concept of validity, and learned how to use the forms of arguments to determine their validity and invalidity. In this lesson we will we bring these two topics together and learn to use truth-tables to test arguments for validity.

As we already know, an argument is valid if and only if it is logically impossible for the conclusion of the argument to be false as the same time that all premises of that argument are true. And if this is the case, then when we set out a valid argument in the form of a truth-table we will find that there are no “TTF lines”: i.e. that there is no line upon which the truth-value of each premise is T and the truth value of the conclusion is F.

An argument is invalid if and only if it is possible for the conclusion of that argument to be false at the same time that the premises of that argument is true. And if this is the case, then when we set out an invalid argument in the form of a truth-table, we will find that there at least one horizontal row upon which the truth-value for each of the premises is true, while the truth-value for the conclusion is F.

#### Example 1: Affirming the Consequent

Consider the following affirming the consequent argument:

 (P → Q) Q P

Now, let’s set this argument out horizontally as a truth-table:

 (P → Q) Q ⊧ P T T T T T 1 T F F F T 2 F T T T F 3 F T F F F 4

As we can see, there is one line (line 3) upon which the two premises of the argument are both true, but the conclusion of the argument is false. As there is a “TTF line” in the truth-table, it must be possible for the premises to be true and the conclusion false at the same time. Thus, we can see from the truth-table that the argument is invalid.

#### Example 2: Modus Ponens

But how do things look when we are dealing with a valid argument? Consider the following modus ponens argument:

 (P → Q) P Q

Now, let’s set this argument out horizontally as a truth-table:

 (P → Q) P ⊧ Q T T T T T 1 T F F T F 2 F T T F T 3 F T F F F 4

As we can see, there is no line on this truth-table upon which both premises are true and the conclusion is false. The only line upon which both premises are true is line 1, and on line 1 the conclusion is also true.

Now let’s consider the hypothetical syllogism form:

 (P → Q) (Q → R) (P → R)

This is how the argument looks set out in a truth-table:

 (P → Q) (Q → R) ⊧ (P → R) T T T T T T T T T 1 T T T T F F T F F 2 T F F F T T T T T 3 F T T T T T F T T 4 T F F F T F T F F 5 F T T T F F F T F 6 F T F F T T F T T 7 F T F F T F F T F 8

Once you get over the fact that this truth-table has eight lines instead of the usual four (eight lines are needed to represent all the possible combinations of truth-values for the three simple sentences contained in the argument) you notice that there are four lines upon which both premises are true (lines 1, 4, 7 & 8) and that the conclusion is also true on all four of these line. This means that the argument is valid.

But now consider the following variation on hypothetical syllogism:

 (P → Q) (Q → R) ⊧ (R → P) T T T T T T T T T 1 T T T T F F F T T 2 T F F F T T T T T 3 F T T T T T T F F 4 T F F F T F F T T 5 F T T T F F F T F 6 F T F F T T T F F 7 F T F F T F F T F 8

As with hypothetical syllogism, there are four lines upon which both premises are true (lines 1, 4, 7 & 8). However, unlike the situation with hypothetical syllogism, on two of these lines the conclusion of the argument is false on two of these lines. This means that the argument is invalid.

The video contains more explanation and examples.

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