Day 7, Mat385

The Zooms continue

A menu of our past zooms...

• 10: I dinged several of you for not explaining. Using too many words is better than not enough.
  1. My favorite: Domain -- animals; Q(x,y) -- "x eats y" [Everything eats something, but nothing eats everything.]

  2. You may not "split your domain". I had several nice examples, but you're not allowed to say, in the domain of integers, that x is negative and y is positive.

     You'd have to add some predicates, which say that N(x) and P(y). x and y themselves must just be arbitrary elements of the domain.

     Another example was a domain of shapes, where the student then said x is a square, and y is a rectangle. Not allowed; x and y are shapes, and you can introduce predicates which serve to dictate their shapes, such as \[ S(x) \land R(y) \rightarrow .... something! \]

  3. One student said to me that if Q(x,y) is (x likes y), then, while it might be true that for anyone y in the world it's true that someone likes y, it's not necessarily true that there's someone who likes every person!
  4. The mistake is in the universal generalization (step 4). In the example above, for a particular y we've chosen the x who likes y; but we can't from that claim that this particular x likes anyone....

     y becomes very special in step 3; in my example (Domain: integers; Q(x,y) as "x+y=0") you can think of it as y=-a!

     Contrast this with #23.

• 13: This says "there is some x and some y such that P(x,y)"; clearly "there is some y and some x such that P(x,y)". But to show it....
  1. Your existential instantiations have to be from the outside in: \[ 1. (\exists x)(\exists y)P(x,y) \] So \[ 2. (\exists y)P(m,y) \] but not \[ 2. (\exists x)P(x,m) \] To this point in the course, we haven't had a rule about doing instantiations within quantified predicates. We have a rule that says that \[ (\exists x)P(x) \rightarrow P(a) \]

     By the way, with the advent of the proof of this exercise, we do have a rule about doing existential instantiations within one type of existentially quantified predicate! If your proof sequence were this,

     \[ \begin{array}{lc} {1. (\exists x)(\exists y)P(x,y)}&{hyp}\cr {2. (\exists y)P(a,y)}&{1, ei}\cr {3. P(a,b)}&{2, ei}\cr {4. (\exists x)P(x,b)}&{3, eg}\cr {5. (\exists y)(\exists x)P(x,y)}&{4, eg} \end{array} \]

     Then you can also conclude that \[ (\exists x)\left[(\exists y)P(x,y)\right] \rightarrow (\exists x)P(x,b) \]

     So you can instantiate within a quantified predicate, in this particular case.

     We keep creating new rules, new tools by proving theorems. But the big upshot of this exercise is that it's legal to interchange the order of two existentials -- they "commute". Note that the point of exercises 10 and 23 is that the two different quantifiers do not commute!

• 23: this is the converse of 10. And it happens to be valid: if there's someone who loves everyone, then certainly for any person there's someone who likes them.

On the Colouring of Maps
Author(s): Professor Cayley
Source: Proceedings of the Royal Geographical Society and Monthly Record of
Geography, New Monthly Series, Vol. 1, No. 4 (Apr., 1879), pp. 259-261