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Is this an example of Denying the Antecedent?

Harry is not an inexperienced driver, and thus must not be irrational, seeing as all inexperienced drivers are irrational.

At first, I thought it was denying the antecedent. Perhaps the repetitious 'NOT' has confused me into thinking "if not q then not p". Otherwise, I am more inclined to opt for undistributed middle because of the larger group of all drivers regardless of experience.

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    At first, I thought it was denying the antecedent. Perhaps the repetitious 'NOT' has confused me into thinking "if not q then not p". Otherwise, I am more inclined to opt for undistributed middle because of the larger group of all drivers regardless of experience. – Carpenoctem Jan 5 at 17:40
  • Correct........ – Mauro ALLEGRANZA Jan 5 at 18:11
  • @MauroALLEGRANZA Are you saying 'correct' to undistributed middle? – Carpenoctem Jan 5 at 18:58
  • I mean it is Denying the antecedent : "All inexperienced drivers are irrational" is "for all x, if x is inexperienced, then x is irrational" and the antecedent is "x is inexperienced". – Mauro ALLEGRANZA Jan 5 at 20:25
  • The argument is invalid because it tells us nothing about how inexperienced driving and irrationality are related. Specifically it does not tell us that all and only inexperienced drivers are irrational. Since the argument gives us no reason to suppose that only inexperienced drivers are irrational, it leaves open that other classes are also irrational and that Harry belongs to one or more such classes. It isn't a case of undistributed middle or of denying the consequent. By process of elimination, then, it's denying the antecedent. Mauro is right. A formalisation would be good. – Geoffrey Thomas Jan 5 at 21:03
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  1. All inexperienced drivers are irrational people.

  2. Harry is not an inexperienced driver.

  3. Thus: Harry is not an irrational person.

The form is AEE in the first figure (All M are P; No S are M; thus, No S are P). The syllogism is invalid. The problem is illicit major; the major term (irrational people) is distributed in the conclusion but not in the premises.

  • Mark: Thanks for the form and mention of illicit major. If you had to answer either i) undistributed middle or 2) idenying the antecedent, which one would you favor more? – Carpenoctem Jan 6 at 1:01
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    @Carpenoctem 1. The middle term (inexperienced driver) is distributed in both premises. 2. The minor premise (2) denies the antecedent of the major premise (1). So I would favor the second choice. – Mark Andrews Jan 6 at 2:02
  • @Mark. I agree the middle term is distributed in both premises - as I found when I recast it, and as you show. That's why I switched to denying the antecedent. But I need to set out the logical form of denying the antecedent and graft the example on to it. Best - Geoffrey. – Geoffrey Thomas Jan 6 at 10:13
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It is a case of denying the antecedent

Let's get the informalities over first.

The argument is invalid because it tells us nothing about how inexperienced driving and irrationality are related. Specifically it does not tell us that all and only inexperienced drivers are irrational. Since the argument gives us no reason to suppose that only inexperienced drivers are irrational, it leaves open that other groups are also irrational and that Harry belongs to one or more such groups.

Denying the antecedent

p → q

not-p


not-q

In other words :

  1. For all x, if x is an inexperienced driver then x is irrational.

  2. x [Harry] is not an inexperienced driver.


  3. x is not-irrational

What's wrong here is that in the second premise the antecedent of the first premise [x is an inexperienced driver] is denied instead of the consequent [x is irrational].

Put it this way. In a deductively valid argument the conclusion cannot be false if the premises are true. But here the premises 1. & 2., could both be true and yet the conclusion, 3., false.

It might indeed be true that 'For all x, if x is an inexperienced driver then x is irrational' and 'x [Harry] is not an inexperienced driver' but the conclusion, 'x is not-irrational' could still be false because x, here = Harry, could be irrational by virtue of his being psychotic or neurotic - a condition which has nothing to do with his being a driver, experienced or otherwise.

The original question scouted the possibility of undistributed middle. Let's dispose of that.

Undistributed middle

In a very compressed nutshell, the middle term is the term which appears in both premises but not in the conclusion. Here the middle term is 'inexperienced driver' and it is properly 'distributed'; it occurs in both premises and is absent from the conclusion. So the middle term is distributed and this is not a case of undistributed middle.

  • Geoffrey: Please see my query further up page where you have selected denying the antecedent rather than undistributed middle. – Carpenoctem Jan 6 at 0:53
  • @Carpenoctem. I will work on the answer. The case is not one of undistributed middle because the middle term both occurs in the two premises and does not occur in the conclusion. I need to fix the logical form of denying the antecedent and map the example on to it. Give me just a while. Best - Geoffrey – Geoffrey Thomas Jan 6 at 10:22
  • @Carpenoctem. I have now explained why this is a case of denying the antecedent. I have set out the logical form of D/A and reframed your argument in terms of it. I have also shown in the same way why U/M does not apply. Hope this helps. Best : GLT – Geoffrey Thomas Jan 6 at 19:18
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At first glance, I did not notice that it is not as valid as it looks like a Modus Ponens using common sense, let us translate it in Propositional Logic (no need for predicate logic).

"all inexperienced drivers are irrational", can be translated as : "If not-experienced then not-rational" .. So : (~E ⊃ ~R)

Now the second sentence :

"Harry is not an inexperienced driver", can be translated not-not-experienced , therefore : ~~E

And the conclusion :

"Thus must not be irrational", can be translated : "not-not-rational", so ∴~~R

The argument would look like :

  1. (~E ⊃ ~R)
  2. ~~E
  3. ∴~~R

And can be simplified to :

  1. (~E ⊃ ~R)
  2. E
  3. ∴R

And yes, it is Denying the antecedent fallacy., therefore invalid.

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