Introduction to Philosophy

Course Home

Thursday, 09-14-17: Logic II

Readings:

Handouts:

Synopsis:

Today we got back Quiz 02, which we took on Tuesday, 9/12. The results were, sadly but not surprisingly, uninspiring. I say sadly, because it would be nice to see some improvement, not surprisingly, though, because these puzzles do increase in difficulty. For better or worse, however, we have to set them aside to concentrate our efforts on learning the techniques we're studying in class. Before getting on to that, let me take a moment to spell out the solution in greater detail.

Recall the puzzle:

The Apprentice was delighted with Abercrombie's reasoning and informed him that he could meet the Sorcerer.

"He is now upstairs in the tower conferring with the island Astrologer," said the Apprentice. "You may go up and interview them if you like, but please knock before entering."

The anthropologist went upstairs, knocked on the door, and was bidden to enter. When he did, he saw two very curious individuals, one wearing a green conical hat and the other a blue one. He could not tell from their appearance which was the Astrologer and which was the Sorcerer. After introducing himself he asked, "Is the Sorcerer a knight?"

The one in the blue hat answered the question (he answered either yes or no), and the anthropologist was able to deduce which was the Sorcerer.

Which one was the Sorcerer?

We must conclude from this that the man in the green hat was the Sorcerer. How?

It's always a good idea to draw pictures, just as it's always a good idea to list what you do know and list what you don't know:

What We Know

  • The Sorcerer is in the room with the Astrologer.
  • Knights always tell the truth while Knaves always lie.
  • Abercrombie is able to figure out whether Blue Hat or Green Hat is the Sorcerer from Blue Hat's answer to his question, "Is the Sorcerer a knight?"

What We Don't Know

  • Whether Blue Hat answered "yes" or "no".
  • Whether Blue Hat or Green Hat is a knight or a knave.
  • Whether the Sorcerer or the Astrologer is a knight or a knave.
  • Whether Blue Hat or Green Hat is the Sorcerer.

Now as much as possible it helps to be systematic. That is, let us fix what we don't know by making assumptions to see where they lead us. Learning how to do this is one of the keys to mastering successful analysis, by the way.

Suppose Blue Hat answered "yes".

Then either Blue Hat is a Knight or Blue Hat is a Knave.

Suppose Blue Hat is a Knight.

Then he is truthfully asserting that the Sorcerer is a knight, but since Green Hat could also be a knight, there is no way to tell which one is the Sorcerer if Blue Hat is a Knight and answers "yes". Yet Abercrombie was able to figure out whether Blue Hat or Green Hat is the Sorcerer!

So suppose Blue Hat is a Knave instead.

Then he is lying when he says the Sorcerer is a knight, hence the Sorcerer is not a knight--he is a knave. Blue Hat is in this case a knave, but Green Hat could be a knave too. Yet once again we remind ourselves that Abercrombie was able to figure out whether Blue Hat or Green Hat is the Sorcerer!

Hence if Blue Hat answers "yes", then it doesn't matter whether he's a knight or a knave. Either way, Abercrombie would not have been able to figure out whether Blue Hat or Green Hat is the Sorcerer.

Hence Blue Hat could not have answered "yes"; he must have answered "no"! We've come a long way since our initial bewilderment.

We know, in particular, that Blue Hat must have answered "no". What follows from a "no" answer? Once again, we have to explore the alternatives of whether Blue Hat is a Knight or a Knave:

Suppose Blue Hat is a Knight.

Then he is truthfully denying that the Sorcerer is a Knight, in which case the Sorcerer is a Knave. But if the Sorcerer is a Knave and Blue Hat is a Knight, then Green Hat must be the Sorcerer.

Suppose Blue Hat is a Knave.

Then he is falsely denying that the Sorcerer is a Knight, in which case the Sorcerer is a Knight. But if the Sorcerer is a Knight and Blue Hat is a Knave, then Green Hat must be the Sorcerer.

Hence since Blue Hat answered "no", it doesn't matter whether Blue Hat is a Knight or a Knave: Either way, Green Hat must be the Sorcerer!

So Green Hat is the Sorcerer, and we are able to deduce that Green Hat is the sorcerer only from the little bits of information we were given and without knowing whether Blue Hat or Green Hat is a knight or a knave.

Pretty cool, eh?

Now, obviously you would not be expected to provide quite so comprehensive an explanation as I provide above, but my goal is to be sure everyone can follow the analysis in solving the puzzle.

Some of the responses Hannah got back were charming: Blue Hat must be the Sorcerer because sorcerers wear blue hats; Green Hat is the Sorcerer because people who speak first are always liars; and so on. Obviously many of you were grasping at straws on this one.

I want to find out who is following along with these synopses. So just for you, you diligent reader, you who so are committed to your education, I'm offering this extra credit puzzle. You'll notice that there are only two possible grades for it. If you get the solution and can explain it, you get a full 20 points; if not, you still get 15 points. So if all you do is draw a pretty picture of knights and knaves, you get 15 points extra credit! Such a deal. Turn this into to me at the start of class Tuesday, 9/19. Do me a favor, though, and keep this to yourself. Just let those from class who find the opportunity earn the points. Does that make me evil?

We began today by putting the Zed-10 group on the spot. It was admittedly chaotic. I gave them the basic points we were to cover today, an example problem to work, and told them they were to take over lecture. I grant, it was asking much. Nevertheless, the key points were, eventually, correctly made. In various places I was even impressed by their work. Still, there was much room for improvement, and I leave it to future Zed-10 groups to figure out how to make that happen.

Let me rehearse the key points briefly here to be sure everyone has got it:

  • A logic, we said last time, is a language on which the relationship of entailment has been defined. Also last time, we gave the syntax and semantics for a simple--yet nevertheless interesting--logic, the Propositional Calculus (PC). Having good notes on the syntax and semantics is crucial, by the way. We do not yet have a logic, however, inasmuch as we haven't defined entailment on PC. To lay the groundwork, last time I gave a general definition of entailment we will use in defining entailment on PC--general, that is, in the sense that it is not defined for any specific langauge. Thus,
  • The General Definition of Entailment: In any language, a set of propositions of the language entails a proposition of the language IFF it is impossible for the entailed proposition to be false while all the entailing propositions are true.
  • Today we gave the first of three definitions of entailment for PC in terms of truth tables. Thus,
  • A set of PC propositions entails a PC proposition IFF the entailed PC proposition is true on every row the entailing PC propositions are all true in a standard truth table. Equivalently,
  • A set of PC propositions entails a PC proposition IFF on no row (of a standard truth table) is the entailed PC proposition false and all the entailing PC propositions true.
  • Since each row of a standard truth table represents a possible combination of truth value assignments to the constitutive sentence letters, and since the standard truth table definitively enumerates in all of its rows all of the possible combinations of truth value assignments, we immediately see that, if on no row is the entailed proposition false while all the entailing propositions are true, it is therefore impossible for the entailed proposition to be false when all the entailing proposition are true.
  • The upshot is that we have i) defined the entailment relation on PC, so PC is now a logic, and ii) we have discovered in the process a method for determining entailments of PC.

We then split up into groups and worked assigned problems--constructing, that is, truth tables to demonstrate entailments from the list of them I gave you last time (and linked above). As I said in class, the scholar would not stop there! No indeed, the scholar would spend part of their weekend mastering the method of Truth Tables by working out all the rest of the entailments given on that handout.

Next time I'll define a Theorem of PC as a kind of limiting case of entailment and we'll proceed to apply the method of Truth Tables to demonstrate the theorems from the handout.

Let me close by saying that I was very proud of all of you who have stuck it out thus far. Now that we have a established a core class of committed scholars, I very much look forward to exploring all this material with you.