An Example of

Symbolic Logic With Truth Tables

We covered the basics of symbolic logic in the last post. Now let’s put those skills to use by solving a symbolic logic statement.

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Generating a Truth Table for (A ∧ ~B) → (C ∨ D)

The key to solving this problem is to break it down into it’s simplest components and take it one piece at a time.

Step 1: We have 4 variables, so we need 4 columns. We’ll also need 16 rows since their are two options for each.

The easiest way to lay out the table is to use an alternating T/F pattern.

In the first column fill in the first half with T and second half with F. Then alternate T/F every 8÷2, or 4, in the second column. Keep the pattern going by alternating T/F every 2 in the third column and every other space in the final column. This covers all possible permutations.

Step 2: We need ~B instead of B, so flip all the truth values in column B. If you’re using a pencil, you can erase column B and simply replace it with ~B.

Step 3: Next we need to compute (A ∧ ~B) and (C ∨ D). We’ll add two more columns onto our table for these computations.

For logical conjunction, i.e. the AND operator, we need both A and ~B to be True to result in True. For OR, we only need one of either C or D to be True to result in True. Use this knowledge to fill in the last two columns.

Step 4: This is the last step! Add one more column to the end for our final computation: (A ∧ ~B) → (C ∨ D).

We know that there is only one case in which implication (→) is false, and that’s when T → F. So check the (A ∧ ~B) and (C ∨ D) columns for this case. There is only one row where this happens in our table. Mark that result as F, and the rest as T.

As messy as this may seem at a glance, it is just a simple application of the definitions presented in the last post. Just remember to always take it one little piece at a time.

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