Source: 7 minutes 22 seconds juncture, Lecture 12-3 (transcription), ... How to Reason and Argue, by Prof Ram Neta PhD (U Pittsbugh; in Philosophy)

For want of brevity, I rewrite the noun phrase 'Breaking the law' as 'lawbreaking'.

[Premise 1:] Lawbreaking is almost always wrong.

[Premise 2:] Double parking is lawbreaking.

[Conclusion:] Double parking is almost always wrong.

Can we refute this argument by means of counterexample? Well, there's no counterexample to premise 2. Premise 2 is simply true, double parking is lawbreaking. [...]

But [...] Is there a counterexample to [...] [premise 1] ? No. [...] Now, how do you produce a counterexample to a claim of the form almost always? Well, the answer is you don't.
Because even if you [...] [exemplify] a case where lawbreaking is NOT wrong, that still doesn't show that  [...]  [Premise 1 is] false, [...]. Maybe lawbreaking is almost always wrong,
but just not in the case that you produced. So, you canNOT
[Caution: The transcript errs here; it states 'can', contrary to the audio]
produce a counterexample to generalization of the form 'lawbreaking is almost always wrong'. That generalization might be false, but you can show that it's false by showing a counterexample.

Okay. So, we cannot refute this argument by means of counterexample. That's not to say that this is a good argument. In fact, this third argument is not a good argument, but we can't show that it's not a good argument by using a counterexample. So, sometimes counterexample can succeed in refuting an argument and sometimes it can't. It depends on whether the argument contains a generalization to the effect that something always happens or something is true in all cases.

What's wrong with the argument above? Premise 2 equates 'double parking' to 'lawbreaking'. This equalization justifies the substitution of one for the other in Premise 1. Now containing 'double parking', this rewritten Premise 1 produces the conclusion.

  • 2
    Ask Clinton. It comes down to what the meaning of 'is', is. In this case 'is' does not represent equality, but inclusion. The 'equation' you are relying upon does not exist, only a containment. Mammals are almost never lay eggs. A platypus is a mammal. Platypuses almost never lay eggs? Well then the poor things are destined for extinction. – jobermark May 28 '15 at 18:27

The fallacy here is that D is not L (by which I mean D is not equivalent to L). L is general whereas D is particular.


  • P1: Canids (Bears) are almost always viscous killers.
  • P2: The tamed circus bear is a canid.
  • C: Therefore, the tamed circus bear is almost always a viscous killer.

This is not valid. You are taking a general set with its own existential characteristics and applying it to an individual in that set which cannot logically be done. The only characteristics that you can say that each individual of the set have are the characteristics defined by the set. This doesn't inherently say anything about the set itself.

For another example, take the set of all positive, whole numbers from 1 to 100. The value is 5050:

  • P1: The individuals in the set of all positive whole numbers from 1 to 100 are almost all double-digit numbers.
  • P2: 100 is an individual in the set of all positive whole numbers from 1 to 100.
  • Therefore, 100 is almost always a double-digit number.

Do you see how this cannot be done? I'm by no means a logician, so I'd appreciate it if someone can help me understand WHY this is the case.


Although a counter-example for the given concrete argument is difficult, one can easily create a counter-example that shows that the general reasoning is invalid.

Let's postulate a brand of playing balls, called JJ. This company makes rubber balls, which are red, and lead balls, which are black. They sell millions of rubber balls each year, but only a handful of lead balls. Those who buy a lead ball generally also buy a large number of rubber balls.

  • Premise 1. A JJ ball is almost always red.
  • Premise 2. A lead JJ ball is a JJ ball.
  • Conclusion. A lead JJ ball is almost always red [wrong!].



That most lawbreaking actions are wrong, leaves room for some lawbreaking actions to be not wrong. What if double parking is precisely those actions for which lawbreaking is not wrong?

It is important to see that premise 2 does not talk about equality or equivalence. A cow is an animal, yet an animal is not always a cow.

Or, in math...

Consider the set B of lawbreaking actions, and the subset W ⊂ B of wrong lawbreaking actions. We know that |W| ≈ |B| (the number of elements in W is approximately equal to the number of elements in B) by premise 1.

Now consider the set DP of double parking actions. We know DP ⊂ B by premise 2. Also consider the set WDP = DP ∩ W of wrong double parking actions (all elements that are in both W and DP).

The conclusion now says that |WDP| ≈ |DP|. However, this does not follow from what we know.

As a toy example, take B = {1,2,3,...,10} and W = {1,2,3,...,9}. Clearly, most elements in B are in W, so premise 1 is fulfilled. Now suppose DP = {10}. We then have WDP = ∅ (empty set). Obviously the conclusion, most DP are in W, is incorrect: no element from DP is in W. Intuitively, this is because DP contains precisely those elements from B that are not in W.

  • Thanks, but in your last sentence, did you mean most DP are in WDP, instead of most DP are in W`? – Greek - Area 51 Proposal Jun 22 '15 at 20:22
  • @LePressentiment the mistake was in the second part of that sentence :) my apologies. – Keelan Jun 22 '15 at 20:37

Convicts are almost always guilty. Wrongfully convicted people are convicts.

Therefore, wrongfully convicted people are almost always guilty [wrong!]

  • 1
    This seems to be no formally different than the argument suggested by Cheers. But without much explanation surrounding it, it doesn't add much as an answer to what we've already got. – virmaior Jun 23 '15 at 12:17

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