32.3: Negation Manipulation

As mentioned above, formal logic is especially useful when we are dealing with very complicated claims. There isn’t a lot that we can do to manage how complex symbolized sentence can get, but one thing we can do is manipulate sentences that contain negations to make them a bit more manageable. In this section, we will look at some rules that will help simplify claims. Simplified claims will be easier to work with in the next chapter, when we start to use tables to test our sentences in various ways.

The easiest and most intuitive way to simplify a claim is to reduce the number of negations. Sometimes, you will find yourself with a double negation. As you might expect, we can take a double negative and cancel them out. So, ~~A can be turned into A. The ‘~’s may be completely removed any time there are an even number of ‘~’’s next to each other. So, ~~~~~~A can be turned into A. Odd numbers of ‘~’’s can be reduced to just one. So, ~~~A becomes ~A and ~~~~A also becomes A. This process is referred to as negation elimination. You won’t often find yourself with a bunch of negations, but it can happen, and when it does, it’s helpful to be able to get rid of them.

Elimination is not the only way to manipulate negation symbols. We may also move them through parenthesis. When you move a negation symbol inside a set of parentheses for ‘&’ and ‘v’ statements, you negate the terms inside of the parenthesis and change the logical operator. So:

• ~(A & B) becomes ~A v ~ B

Likewise:

• ~(A v B) becomes ~A & ~B

You may also move negations outside of parentheses through a similar process. To remove negations from parentheses, remove a negation symbol from each term and change the logical operator. So:

• ~A v ~B can move back to ~(A & B) and
• ~A & ~B can move back to ~(A v B)

If this seems strange, in the next chapter we will learn how to test claims to see if they are logically equivalent using truth tables. At that time, you will be able to use tables to prove that these claims are saying the same thing (logical equivalence), but for now let’s just think it through. When someone says:

• “I don’t like Thai or Indian food.”

they are actually saying:

• “I don’t like Thai food” and “I don’t like Indian food”.

Likewise, when someone says:

• “I’m not going to date both Amy and Carl.”

They mean that they aren’t going to date both Amy or Carl (but maybe they will date one of them). Things get more complicated when it comes to moving negations for conditionals and biconditionals. For conditionals, you can move a negation inside a set of parentheses by negating the first term and changing the operator to an ‘&’. So:

• ~(P → Q) becomes P & ~Q

Thinking back to Chapter 3, you should recall that a conditional says that if the antecedent is true then the consequent is true. If someone is negating a conditional, then they must be telling us that when the antecedent is true the conditional is false, thus: P & ~Q. You can also remove negations in the reverse way. So:

• P & ~Q can become ~(P→Q)

For now, it’s ok for you to just accept that this rule works, and you will be able to use the test for logical equivalence in the next chapter to prove it.

Biconditionals are even more complicated. A negated biconditional:

• ~(P→ Q), becomes (~P & Q) v (P & ~Q)

Remember that a biconditional is really saying these things happen together and never alone. So, if the biconditional is being negated, it means P or Q are happening alone. Again, this can be reversed. So:

• (~P & Q) v (P & ~Q) can be simplified to ~(P ← → Q)

In all honesty, the biconditional manipulation rule doesn’t come up all that often, but you will be thankful when you can turn 5 logical operators, two sets of variables, and two sets of parentheses into two variables, two operators, and one set of parentheses. Again, if the reason doesn’t make sense now that’s fine. In the next chapter you will learn how to prove it.

Exercises in Negation Manipulation

Simplify the following sentences by using negation manipulation.

1. ~~A v ~~B
2. ~ (~A v ~B)
3. ~ (~A & ~B)
4. ~ (~~A v B)
5. ~ (~A & B)
6. ~~~ [A → (~~B v A)]
7. (~P & Q) v (P & ~Q)
8. ~A v (~B v ~C)
9. ~A → (~B ← →~C)
10. (A v B) & ~C
11. ~A & [(~B v C) & D]
12. ~[(~P & Q) v (P & ~Q)]
13. ~~~A → (~~~B & ~C)
14. ~ A & [~B &(~C & D)]
15. [(A v ~~~B) v C] & ~~D
16. ~A v (~~B & ~~C)
17. A → (B → C)
18. B ← → ~~C
19. (~A v B)
20. ~~~[~~~~A → (~~~B ← →~~~~~C)

Answers to Selected Exercises

6. ~~~ [A → (~~B v A)]

The first step is always to remove any pairs of negations (remember that odd numbers of negations adjacent to each other can always be reduced down to one, and even numbers can be reduced to zero). This sentence doesn’t reduce any further, so the final answer is: ~ [A → (B v A)].

10. (A v B) & ~C

This one is already as simple as it can be.

13. ~~~A → (~~~B & ~C)

First, remove any pairs of negations to get: ~A → (~B & ~C). Then, pull the negation out of the parentheses: ~A → ~(B v C). There is no rule for what to do when a conditional looks like this, so you’re done.

14. ~ A & [~B &(~C & D)]

There are no pairs of negations, so we can skip that step. Looking at the problem it might not be clear, but we can make things simpler. The key is to use the rule for pulling a negation out of an ‘and’. This gets us: ~ A & [~B &~(C v ~D)]. Notice that since D was not negated before, we have to add a ~ to it. Now, we can do the part in the brackets: ~ A & ~[B v (C v ~D)]. We can take it one step further by adding another set of ( ) around what we have and pulling the ~ out: ~(A v [B v (C v ~D)]). There is nothing left to pull out, so we’re done.

16. ~A v (~~B & ~~C)

The first thing to do is remover the pairs of negations: ~A v (B & C). With that done, the sentence is as simple as it can be. You could change the ‘v’ to an ‘&’ by adding ( ) and pulling the negation out, but that would make it more complicated.

17. A → (B → C)

We don’t have a rule to deal with this case so there is nothing to do.