Truth Tables and Validity-2

Let's test a slightly more complex argument for validity using the truth table method.  Consider the argument:

(P Q), (~R ~Q)  therefore, (P R)

Since there are three variables, we need an 8 row truth table.

(Step 1)  Set up the basic truth table:

 1 2 3 4 5 6 7 8 9 10 11 12 13 14 P Q R (P → Q) (~ R → ~ Q) ∴ (P → R) T T T T T T T T T T T ⊥ T T ⊥ T T ⊥ T ⊥ T T ⊥ T ⊥ T T T ⊥ ⊥ T ⊥ ⊥ ⊥ T ⊥ ⊥ T T ⊥ T T T ⊥ T ⊥ T ⊥ ⊥ T ⊥ T ⊥ ⊥ ⊥ ⊥ T ⊥ ⊥ T ⊥ ⊥ T ⊥ ⊥ ⊥ ⊥ ⊥ ⊥ ⊥ ⊥ ⊥

Note that the important columns here are columns 5, 9 (the premises) and 13.

(Step 2)  Fill in Columns 7 and 10, as the negations have the smallest scope and those columns are components of Column 9.

 1 2 3 4 5 6 7 8 9 10 11 12 13 14 P Q R (P → Q) (~ R → ~ Q) ∴ (P → R) T T T T T ⊥ T ⊥ T T T T T ⊥ T T T ⊥ ⊥ T T ⊥ T ⊥ T T ⊥ ⊥ T T ⊥ T T T ⊥ ⊥ T ⊥ T ⊥ T ⊥ T ⊥ ⊥ T T ⊥ T ⊥ T ⊥ T ⊥ T ⊥ T ⊥ ⊥ T T ⊥ ⊥ T ⊥ ⊥ ⊥ ⊥ T ⊥ ⊥ ⊥ T T ⊥ ⊥ T ⊥ ⊥ ⊥ ⊥ ⊥ T ⊥ T ⊥ ⊥ ⊥

Now complete columns 5, 8 and 13.  For column 8 be sure to remember that column 7 is the antecedent and column 10 is the consequent.

 1 2 3 4 5 6 7 8 9 10 11 12 13 14 P Q R (P → Q) (~ R → ~ Q) ∴ (P → R) T T T T T T ⊥ T T ⊥ T T T T T T ⊥ T T T T ⊥ ⊥ ⊥ T T ⊥ ⊥ T ⊥ T T ⊥ ⊥ ⊥ T T T ⊥ T T T T ⊥ ⊥ T ⊥ ⊥ T ⊥ T T ⊥ T ⊥ ⊥ ⊥ T T ⊥ T T ⊥ T T ⊥ T ⊥ T T ⊥ T ⊥ ⊥ T T T ⊥ ⊥ ⊥ T ⊥ T ⊥ ⊥ ⊥ T ⊥ T ⊥ ⊥ T T T ⊥ ⊥ T T ⊥ ⊥ ⊥ ⊥ T ⊥ T ⊥ T T ⊥ ⊥ T ⊥

Since there is no row in which both column 5 and column 9 are true while column 13 is false, that is, there is no invalidating row, the argument is valid.

Finally, let's consider the following argument:

(P  Q),  (R  S),  (~P ~R)     therefore (~Q  ~S)

(Step 1)  Construct the basic truth table for the argument.  Since there are 4 variables, there will be 16  rows.

 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 P Q R S (P → Q) (R → S) (~ P ● ~ R) (~ Q ● ~ S) T T T T T T T T T T T T T T T ⊥ T T T ⊥ T T T ⊥ T T ⊥ T T T ⊥ T T ⊥ T T T T ⊥ ⊥ T T ⊥ ⊥ T ⊥ T ⊥ T ⊥ T T T ⊥ T T T T ⊥ T T ⊥ T ⊥ T ⊥ T ⊥ T T ⊥ ⊥ T ⊥ ⊥ T T ⊥ ⊥ T T ⊥ ⊥ T T ⊥ ⊥ ⊥ T ⊥ ⊥ ⊥ T ⊥ ⊥ ⊥ ⊥ T T T ⊥ T T T ⊥ T T T ⊥ T T ⊥ ⊥ T T ⊥ ⊥ T T ⊥ ⊥ T ⊥ T ⊥ T ⊥ T ⊥ ⊥ T T ⊥ T ⊥ ⊥ ⊥ T ⊥ ⊥ ⊥ ⊥ T ⊥ ⊥ ⊥ T T ⊥ ⊥ T T ⊥ T ⊥ T ⊥ ⊥ T ⊥ ⊥ ⊥ T ⊥ ⊥ T ⊥ ⊥ ⊥ ⊥ ⊥ T ⊥ ⊥ ⊥ T ⊥ ⊥ ⊥ T ⊥ ⊥ ⊥ ⊥ ⊥ ⊥ ⊥ ⊥ ⊥ ⊥ ⊥ ⊥

Note that in this truth table the relevant columns are columns 6, 9, 13 (the premises and 18 (the conclusion. If we find ANY row in which 6, 9 and  13 are true, while 18 is false, the argument is NOT Valid.

(Step 2)  Fill in the columns for the operators with the smallest scope--columns 11, 14, 17, and 19.

 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 P Q R S (P → Q) (R → S) (~ P ● ~ R) (~ Q ● ~ S) T T T T T T T T ⊥ T ⊥ T ⊥ T ⊥ T T T T ⊥ T T T ⊥ ⊥ T ⊥ T ⊥ T T ⊥ T T ⊥ T T T ⊥ T ⊥ T T ⊥ ⊥ T ⊥ T T T ⊥ ⊥ T T ⊥ ⊥ ⊥ T T ⊥ ⊥ T T ⊥ T ⊥ T T T ⊥ T T ⊥ T ⊥ T T ⊥ ⊥ T T ⊥ T ⊥ T ⊥ T ⊥ ⊥ T ⊥ T T ⊥ T ⊥ T ⊥ ⊥ T T ⊥ ⊥ T ⊥ T T ⊥ T ⊥ ⊥ T T ⊥ ⊥ ⊥ T ⊥ ⊥ ⊥ ⊥ T T ⊥ T ⊥ T ⊥ ⊥ T T T ⊥ T T T T ⊥ ⊥ T ⊥ T ⊥ T ⊥ T T ⊥ ⊥ T T ⊥ T ⊥ ⊥ T ⊥ T T ⊥ ⊥ T ⊥ T ⊥ T ⊥ T T ⊥ T ⊥ ⊥ T ⊥ T ⊥ T ⊥ ⊥ ⊥ T ⊥ ⊥ T ⊥ T ⊥ ⊥ T T ⊥ ⊥ ⊥ T T ⊥ ⊥ T T T ⊥ ⊥ T T ⊥ ⊥ T ⊥ ⊥ T ⊥ ⊥ ⊥ T ⊥ T ⊥ ⊥ T T ⊥ T ⊥ ⊥ ⊥ ⊥ T ⊥ ⊥ ⊥ T T ⊥ T ⊥ T ⊥ ⊥ T ⊥ ⊥ ⊥ ⊥ ⊥ ⊥ ⊥ ⊥ T ⊥ T ⊥ T ⊥ T ⊥

Now construct the columns for the remaining operators, columns 6, 9, 13, and 18.  Remember, columns 11 and 14 give you the input values for column 13, and columns 16 and 19 give you the input values for column 18.

 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 P Q R S (P → Q) (R → S) (~ P ● ~ R) (~ Q ● ~ S) T T T T T T T T T T ⊥ T ⊥ ⊥ T ⊥ T ⊥ ⊥ T T T T ⊥ T T T T ⊥ ⊥ ⊥ T ⊥ ⊥ T ⊥ T ⊥ T ⊥ T T ⊥ T T T T ⊥ T T ⊥ T ⊥ T ⊥ ⊥ T ⊥ ⊥ T T T ⊥ ⊥ T T T ⊥ T ⊥ ⊥ T ⊥ T ⊥ ⊥ T ⊥ T ⊥ T ⊥ T T T ⊥ ⊥ T T T ⊥ T ⊥ ⊥ T T ⊥ ⊥ ⊥ T T ⊥ T ⊥ T ⊥ ⊥ T ⊥ ⊥ ⊥ T ⊥ ⊥ T T ⊥ T T ⊥ T ⊥ ⊥ T T ⊥ ⊥ ⊥ T T ⊥ T ⊥ T ⊥ T ⊥ ⊥ ⊥ T T ⊥ ⊥ ⊥ T ⊥ ⊥ ⊥ T ⊥ ⊥ T ⊥ T ⊥ T ⊥ T T ⊥ ⊥ T T T ⊥ T T T T T T ⊥ ⊥ ⊥ T ⊥ T ⊥ ⊥ T ⊥ T T ⊥ ⊥ T T T ⊥ ⊥ T ⊥ ⊥ ⊥ T ⊥ T ⊥ T ⊥ ⊥ T ⊥ T ⊥ T T ⊥ T T T ⊥ T T ⊥ ⊥ T ⊥ ⊥ T ⊥ T ⊥ ⊥ ⊥ T T ⊥ T ⊥ T ⊥ T T ⊥ ⊥ T ⊥ T ⊥ ⊥ ⊥ T T ⊥ T ⊥ T T T T ⊥ ⊥ ⊥ T T ⊥ ⊥ ⊥ T ⊥ ⊥ T ⊥ ⊥ T ⊥ T ⊥ ⊥ T ⊥ ⊥ ⊥ T T ⊥ T T ⊥ ⊥ ⊥ ⊥ T ⊥ T ⊥ ⊥ T T T ⊥ T T ⊥ T ⊥ ⊥ ⊥ T ⊥ ⊥ ⊥ ⊥ ⊥ T ⊥ ⊥ T ⊥ T ⊥ T T ⊥ T ⊥ T T ⊥

Since there is at least one row in which all the premises are true and the conclusion is false, this argument is NOT valid.