# What fallacy is it to say all teenagers are bad, therefore all non-teenagers are not bad?

Suppose there is a non-empty subset A of U. Let A' denote the complement of A in U.

What is the name of this logical fallacy?

X is true for A
therefore
not X is true for­­­­­­­­­­­­­­­­­­ A'

For example, suppose U = {all people} and A = {all teenagers} then the logical fallacy is

all teenagers are bad
therefore
all non-teenagers are not bad

• Is there a reason you're using different notations for not X and A' ? Either is the complement of X or A (respectively). Jan 10, 2020 at 15:22
• @Flater If I used the same notation for X what would the superset be? Jan 11, 2020 at 4:01
• Actually, I could just use X ⊆ {bad, not bad}. If I had realized that earlier I might not have had to ask the question! Jan 11, 2020 at 4:08
• The definition of not inherently means that { X , not X } is a complete set. You don't need to define it for every variable, negation is a basic operation. Jan 11, 2020 at 16:43

## 5 Answers

After some thought I realized this is a denying the antecedent fallacy. Put another way we have

If the person is a teenager then they are bad
therefore
If the person is a non teenager then they are not bad

• ‘Denying the antecedent’ correctly identifies the fallacy. ‘All not-bad persons are not teenagers’ is the contrapositive. Apr 16, 2023 at 18:01

To add to @Jon's answer, denying the antecedent often comes up due to confusion with the valid argument form Modus Tollens

((p → q) ∧ ¬q) → ¬p)
or ((if p then q) and not q) then not p)

Which is equivalent to

(if p then q. Therefore, if not q then not p), i.e.

If a person is a teenager, then they are bad
Therefore
If a person is not bad, they are not a teenager

@willross1 currently writing a thesis on logical fallacies, new to this forum. I am confused about your account of modus tollens. I believe it is a simple clerical error in which you have accidentally forgotten a parentheses? There are too many close parentheses per open parentheses. Is this simply a keyboard error or am I too prone to misunderstand MT? {[(p➡️q) & ~q] ➡️ ~q}

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– J D
Jan 16, 2020 at 0:40

Other contributors gave really interesting answers, yet no one mentioned the link to the classical confusion of necessity and sufficiency - the X is necessary for being an element of A, i.e. A ⊆ T, where T = {t ∈ U | X is true for t}. Yet no one guarantees that A = T (property is not a criterion), thus, ¬X is not necessary for being an element of A' (unless A = T is stated), but it sufficient for being an element of A'. It could be demonstrated via the following Euler diagram (sorry for a sloppy picture):

This is a case of modus tollens.

Say,

Q implies bad

P implies teenager

According to your problem,

P implies Q|!P|Therefore, !Q|

This shows that if teenager is bad, non-teenager means not bad.