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Please help to check my understanding of Example 1.2.3 from "How to prove it" by Daniel Velleman.

Determine whether the following arguments are valid.

  • Either John isn’t stupid and he is lazy, or he’s stupid.
  • John is stupid.
  • Therefore, John isn’t lazy.

... we let S stand for the statement “John is stupid” and L stand for “John is lazy.” Then the argument has the form:

  • (¬S ∧ L) ∨ S
  • S
  • ∴ ¬L

The author concludes that John isn't lazy.

Why John cannot be both stupid and lazy here?

Provided truth table:

S L (¬S∧L)∨S S ¬L
F F    F     F  T
F T    T     F  F
T F    T     T  T
T T    T     T  F

The author gives additional clarifications but they are not clear for me.

If you expected the first argument in Example 1.2.3 to turn out to be valid, it’s probably because the first premise confused you. It’s a rather complicated statement, which we represented symbolically with the formula (¬S ∧ L) ∨ S. According to our truth table, this formula is false if S and L are both false, and true otherwise. But notice that this is exactly the same as the truth table for the simpler formula L ∨ S! Because of this, we say that the formulas (¬S ∧ L) ∨ S and L ∨ S are equivalent. Equivalent formulas always have the same truth value no matter what statements the letters in them stand for and no matter what the truth values of those statements are. The equivalence of the premise (¬S ∧ L) ∨ S and the simpler formula L ∨ S may help you understand why the argument is invalid. Translating the formula L ∨ S back into English, we see that the first premise could have been stated more simply as “John is either lazy or stupid (or both).” But from this premise and the second premise (that John is stupid), it clearly doesn’t follow that he’s not lazy, because he might be both stupid and lazy.

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  • $\begingroup$ I think the reason you are confused is that the author's argument is an example of invalid reasoning. $\endgroup$ – ajd Feb 24 '17 at 1:05
  • $\begingroup$ "Why John cannot be both stupid and lazy here?" It can, but in this case, the two premises "Either John isn’t stupid and he is lazy, or he’s stupid" and "John is stupid" are satisfied (i.e true), while the purported conclusion "John isn’t lazy" is not (i.e. is false) : see last row of the t-t. Thus, the conclusion does not follow from the premises, i.e. the argument is invalid. $\endgroup$ – Mauro ALLEGRANZA Feb 24 '17 at 14:57
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You are correct. Polya is asking you to determine whether or not the argument is valid.

If you expected the first argument in Example 1.2.3 to turn out to be valid, it’s probably because the first premise confused you.

Here he is stating that the conclusion of the argument (John is not lazy) is invalid. Furthermore,

But from this premise and the second premise (that John is stupid), it clearly doesn’t follow that he’s not lazy, because he might be both stupid and lazy.

Here Polya is giving the exact concern that you raise -- the statement is true regardless of how lazy John is. Hence the statement

it clearly doesn't follow that he's not lazy

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