How not to teach logic

Some faculty at York think it’s ok to omit parentheses. There’s not much to say other than “wow”.

Take a look at the first and most fundamental formulas of predicate calculus, as summarized on a formula sheet for an intorductory logic course:

The first problem is that the author is employing the over-used triple-bar symbol to denote material bi-implication, as opposed to the much more reasonable and popular ↔ symbol. The second problem is that these formulas are so ambiguous that they can’t even be said to be false.

But, I hear you say, one should perhaps assume that the parentheses are implicit, and moreover that they are left-associating by default. So in fact, in the context of the class, these formulas may well be correct! Such an argument, aside from the fact that it presupposes the instructor is brewing up a pedagogical nightmare of a syllabus, is initally somewhat convincing. Until you read a little further down on the page:

Seriously, just look at it.

To be clear, this isn’t the worst of it. The professor who compiled these notes at least included some parentheses, you know, for when things get super-duper-hyper ambiguous. Unfortunately, a number of other professors teach the same course from the same notes. Only when they copy the notes, they don’t tend to include any parentheses (I’ll scan it if I see it again).

What does this say to students? It says, loudly and clearly:

I’m too busy to worry about whether my course materials are universally confusing or not. In fact, figuring this shit out is really your job. If you can’t figure out what I meant, then I don’t see how you could pass this course anyway.

And to that I would add: “Besides, I don’t give two dips about the material, since it doesn’t have anything to do with the way mathematicians think about mathematics anyway. Why are we studying it? Something about maturity. Ask the department chair.”

Update. Martino points out the following: If you insist on a particular set of rules for operator precedence, and if you then insert enough parentheses to make these ambiguous formulas well-formed, then they will be true regardless of which choices you made along the way!

This doesn’t, of course, make the formulas any less confusing. In some ways it makes it more confusing: that several axioms (some of which are more perverse than others) are being combined into one without comment.

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8 Comments

  1. Joel David Hamkins
    Posted February 10, 2012 at 8:10 pm | Permalink

    Sam, although I agree that the notes are unbelievable, I wonder whether it is wise for you to be outing a colleague?

  2. Samuel Coskey
    Posted February 11, 2012 at 12:41 am | Permalink

    In this case, I am not connected with the individuals and it would not be appropriate for me to approach them in private. One can hope this post will reach them (or their boss), but unfortunately I doubt it.

  3. Micheal Pawliuk
    Posted February 12, 2012 at 5:47 pm | Permalink

    I’m too busy to worry about whether my course materials are universally confusing or not. In fact, figuring this shit out is really your job. If you can’t figure out what I meant, then I don’t see how you could pass this course anyway.

    I laughed hard at that. :)

  4. Carl Mummert
    Posted February 13, 2012 at 11:25 pm | Permalink

    What is the “golden rule” supposed to be saying?

    • Samuel Coskey
      Posted February 13, 2012 at 11:28 pm | Permalink

      Treat others as they would wish to be treated?

      • Carl Mummert
        Posted February 14, 2012 at 8:13 am | Permalink

        I mean in the second displayed excerpt. I don’t even recognize the rule.

    • Samuel Coskey
      Posted February 16, 2012 at 10:29 pm | Permalink

      On closer inspection, I believe this one actually turns out to be false. It is close to a true rule though, which is that the equivalence of $A\wedge B$ with $A$ is itself equivalent to $B$. Or something.

    • Posted February 18, 2012 at 11:27 am | Permalink

      I think it should read $((A \land B) \leftrightarrow A) \equiv (B \leftrightarrow (A \lor B))$. (Reading it as $\inf(a,b) = a \Leftrightarrow b = \sup(a,b)$ makes it easier to understand why it is true.)

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