# Which fallacy is this? “For every A, there is a B. Therefore, for every B there is an A.”

I know that the following reasoning is a logical fallacy, but I don't know how to classify it:

"For all A there is a corresponding B. Therefore for all B there is a corresponding A."

It is not exactly affirming the consequent, I believe, because the above statement includes quantification. Is there a special form of "affirming the consequent" in first order logic?

• It is, as you implicitly suggest, a formal fallacy rather than an informal one, and relatively few formal fallacies have names, because we don't need to name them to identify them, since we can recognize them in first-order logic. – ChristopherE Dec 10 '13 at 17:46
• Although in this case there's a name, "syllogistic error", I believe. This is a logical error, but all too common. – Michael Dec 10 '13 at 18:31
• I think it is called "non sequitur". – Ingo Dec 10 '13 at 20:50
• @Ingo, Well just in the sense that a "non sequitur" is anything that does not follow from something else. – ChristopherE Dec 11 '13 at 0:43
• I'll have to agree with @Ingo. – Tico Dec 18 '13 at 0:47

The general fallacy here is a typical confusion with quantifier order, and I would call it a "quantifier-order fallacy". This is a phenomenon that tends to be revealed much more commonly and clearly in the study of contemporary first-order logic than it is in ancient syllogistic logic, simply because it is quantification and particularly the importance of quantifier order that is made clear and emphasized in first-order logic.

The assertion "for every A there is a corresponding B" is asserting merely that the relation is total, that every A has corresponding B. But the assertion "every B has a corresponding A" is asserting that the relation is surjective.

Students in mathematics are quite commonly confused about this distinction. For example, they commonly conflate what it means to say that a function f:X→Y is one-to-one, or that it is onto Y. For example, suppose that f:X→ Y is a function, which means that for every a∈ X there is a unique b∈ Y such that b=f(a).

• the assertion ∀a∈X ∃b ∈ Y b=f(a) is saying merely again that f is a function.

• but the assertion ∀b∈Y ∃a ∈ X b=f(a), with the quantifiers reversed, is saying that f is an onto function, that every point in Y is hit by the function.

• meanwhile, the related concept that is often confused is the assertion that f is one-to-one, which is ∀a,a'∈X f(a)=f(a') → a=a'.

Perhaps it is fair to say that one can make quite sophisticated and complex mathematical statements with just a few alternations of quantifiers. This distinction is what underlies the difference between continuity of a function and uniform continuity, which was classically confusing to many prominent mathematicians.

Meanwhile, there are abundant natural language examples of the quantifier-order fallacy:

• Every boy was kissed by a girl, versus
• There is a girl who kissed every boy.
• Every child at the circus was supervised by an adult.
• There was an adult that supervised every child at the circus.
• every time the train runs, there is someone who is mugged.
• there is someone who is mugged every time the train runs.

But actually, if one looks at your fallacy, it isn't just the quantifier order that is reversed, but the actual quantifiers themselves are swapped. So some more accurate natural language instances would be:

• Every boy was kissed by a girl, versus
• every girl kissed a boy.
• every child was supervised by an adult, versus
• every adult supervised a child.
• every person at the performance was in a seat, versus
• every seat at the performance was occupied by a person.

This is the same issue that arises in the distinction between

• f:R→R is continuous iff for every x, for every positive ε there is positive δ for every x', if |x-x'|< δ → |f(x)-f(x')|< ε.
• f:R→R is uniformly continuous iff for every positive ε there is positive δ for every x, for every x', if |x-x'|< δ → |f(x)-f(x')|< ε.