Supplement to Chapter 20 Another Form of DeMorgan’s Rule: The DeMorgan Algorithm

DeMorgan’s rule (“DM”) allows us to replace a sentence of the form ~(P & Q) with the corresponding sentence ~ P v ~ Q, and vice versa, that is, it allows us to replace a sentence of the form ~ P v ~ Q with the corresponding sentence ~(P & Q). Of course, DeMorgan’s rule also allows us to replace a sentence of the form ~ (P v Q) with the corresponding sentence ~ P & ~Q, and vice versa. Notice that DM “drives” the tildes inside, that is, in these cases, it shifts the tilde from the outside of the brackets onto the inside letters. At the same time, the wedge changes to the ampersand or the ampersand changes to the wedge, of course.

An algorithm is a very precise, step-by-step procedure for carrying out a specific task. The purpose of this supplement is to present a little three-step algorithm that captures the essential logic of the DeMorgan rule. Using this algorithm, you will be able to make every replacement you would make using the original DeMorgan rule, plus many more. I like to add this algorithm to our proof system, and thus count any use of it as an application of the DM rule. Here is the “DeMorgan Algorithm.”

DeMorgan Algorithm If a formula, or a subformula within a formula, contains an ampersand, or wedge, as its main connective, you may: 1. Change the ampersand to a wedge, or the wedge to an ampersand, as the case may be. 2. Add a tilde to the formula to the immediate left of the ampersand or wedge, and add a tilde to the formula to the immediate right of the ampersand or wedge. 3. Add a tilde to the formula or subformula as a whole. Let us now apply the DM algorithm to the formula (A v B). The process involves three easy steps:

Given: (A v B) 1. First, we change the wedge to an ampersand: (A & B) 2. Next, we add a tilde to each side of the “amp.” So, we add a tilde to the A and we add a tilde to the B: (~ A & ~ B) 3. Finally, we add a tilde to the whole: ~( ~A & ~ B)

Thus, the formula (A v B) is equivalent to, and thus is interchangeable with, the formula ~( ~A & ~B). This means that one may replace the other anywhere in a proof. (This equivalence can easily be confirmed on a truth-table, by the way, if you studied Chapters 14-17.)

Let’s try another. Here are the three steps when we apply the DM algorithm to ~ (A & B):

Given: ~( A & B) Step 1. ~(A v B) (We changed the & to v.) Step 2. ~(~A v ~ B) (We added a tilde to each side of the v.) Step 3. ~ ~ ( ~A v ~ B) (We added a tilde to the whole.) Thus, ~ (A & B) is equivalent to ~ ~ (~A v ~ B). This, too, can easily be confirmed on a truth- table. However, this new formula has way too many tildes! This can be remedied by applying Double Negation. If we apply DN to ~ ~ ( ~A v ~B) we get: ( ~A v ~ B). (We removed the double tildes on the outside of the brackets.) Now we can say:

~(A & B) is equivalent to ~A v ~B.

This means that one formula may replace the other anywhere in a proof. Notice that the DM algorithm, in this case, gives the same result as the original DeMorgan’s rule.

Here are additional examples. In each of the following, the second formula was derived from the first by applying the DM algorithm: 1. ~(A v ~ B) 1. (~ A & ~B) 1. (A & ~B) 2. ~ ~ (~ A & ~ ~B) 2. ~( ~ ~A v ~ ~ B) 2. ~(~A v ~ ~ B) Note: not all the steps are shown, only the first and last step.

However, again, the second formula in each case has too many tildes. But this can be fixed by applying Double Negation. If we apply DN to each one, we get this result: 3. (~ A & B) 3. ~( A v B) 3. ~(~A v B) Less clutter, much nicer. Remember that DeMorgan’s rule applies only to conjunctions and disjunctions--it does not apply to horseshoes and triple bars.

A Correction in the Textbook

In Exercise 20.1, on page 406 of the textbook, in problems 2 and 3, the last step in each proof was actually derived using the DeMorgan algorithm, rather than the original DeMorgan rule. But the DM algorithm is not taught in the chapter. I included the DM algorithm in my previous logic text, The Many Worlds of Logic, but took it out of the new text, Introduction to Logic, in order to cut down on the page count. So that is an error on my part! Sorry about that. To get the last step in problem 2 using the DeMorgan rule, you would need to add a step. You would need to first apply Double Negation to line 9, and get: ~ ~(F v S). After that, the DM rule would allow you to go straight to the conclusion, ~(~F & ~S). The same “move” is required to get the last step of problem 4 by the DM rule in the text as well. As you can see, the algorithm can sometimes save you one step. Good luck with the proofs!