# is "2+3=6, then 2+3+1=7" an example of the fallacy fallacy?

Is "if 2+3=6, then 2+3+1=7" an example of a fallacy fallacy or argumentum ad logicam?

I believe it is an example of the fallacy fallacy, but I'm not able to show it. I only think it is a fallacy because "2+3=6" and "2+3+1=7" are false.

In formal logic, the conditional between two false statements is valid. What if we check the conditional statement using informal logic? Is it an example of a fallacy in informal logic, especially the argumentum ad logicam?

• This is a case of ex falso quodlibet. You can prove any statement if your assumptions are contradictory. Feb 17, 2013 at 17:49
• It's a true material implication. What's the problem? If 2+2=5 then I am the Pope. Jul 25, 2014 at 18:12
• To expand on CodesInChaos and user4894's comments: From the rules of basic arithmetic we have: 2+3=5 and 5=/=6. So 2+3=/=6. Combining this with your initial premise, we obtain the contradiction 2+3=6 and 2+3=/=6 from which any proposition follows, including 2+3+1=7. Jun 10, 2015 at 3:16

The "fallacy fallacy" is that you reject a conclusion because the argument is a fallacy - even though the conclusion may be right. In general form:

1. If P, then Q.
2. P is a fallacious argument.
3. Therefore, Q is false.

You have only given step 1 and step 2; you didn't accept step 3, rejecting the consequent (Q). Since the implication (P->Q) holds (step 1) and P is indeed a fallacious argument (step 2), there is no fallacy. If you had concluded that 2+3+1=7 was wrong based on the antecedent - and not on your own 'mathematical intuition', then it would have been a fallacy.

The implication "If 2+3=6, then 2+3+1=7" is true, since both the antecedent and the consequent are false. You can look at the truth table for this: An example of an argumentum ad logicam is the following:

1. If 2+3=6, then 4+4=8.
2. 2+3=6 is false.
3. Therefore, 4+4=8 is false.

There is nothing wrong with step 1 and 2. The implication (step 1) holds and "2+3=6" is false (step 2). It does not logically follow, however, that "4+4=8" is false (step 3).

Note that the implication in step 1 also holds, but that there is a difference with the previous example I gave. Here the antecedent is false, but the consequent is true (which can lead to the fallacy fallacy). Truth table: (As a side note: a general rule to determine the truth value of an implication: if the value of q≥p, the implication holds (even if q and p are completely unrelated!). If you find this a bit absurd, you're not alone. Many Logicians find it a quite troublesome.)

Or I can give a more absurd example:

1. If New York is in China, then New York is in the U.S.A.
2. New York is not in China.
3. Therefore, New York is not in the U.S.A.

The truth table in this case is the same as in the previous example: Here, the implication in step 1, as absurd as it may sound, is true. This may illustrate the point I made in the previous example: the implication can be quite troublesome at times.

In real-life, this will often occur in a more subtle manner. For instance:

Argument by proponent:

1. Socrates is a sentient being.
2. All humans are sentient beings.
3. Therefore, Socrates is human.

Counter-argument by opponent:

1. You have just affirmed the consequent.
2. You have committed a fallacy; your reasoning was wrong, so Socrates is not human.

While it is of course true that the argument given by the proponent is fallacious, his conclusion was correct. The opponent could point at his flawed reasoning, but he could not reject the conclusion based solely on this flawed reasoning. This is an example of the fallacy fallacy or argumentum ad logicam.

The conclusion drawn is absolutely correct: If 2 + 3 were 6 (which it isn't) then 2 + 3 + 1 would be one more than 2 + 3, so it would be 6 + 1 which is 7 (but of course it isn't). The conclusion is also very obvious and direct.

As "CodesInChaos" said: "This is a case of ex falso quodlibet. You can prove any statement if your assumptions are contradictory". To explain a bit: If 2 + 3 were 6, then we can subtract 5 from each side of the equation and get 0 = 1. We then can multiply both sides by any x, or by any y, and get 0*x = 1*x or 0 = x, and 0*y = 1*y or 0 = y. And since 0 equals any x, and it equals any y, we have x = y, whatever x and y are. And now we can prove anything. For example we prove that Fermat's Last Theorem is actually wrong: It says that a^n + b^n = c^n has no solutions if a, b, c > 0 and n >= 3. But if we let x = a^n + b^n, and y = c^n, then the equation is actually true, for all values of a, b, c and n!