P310 Valid Reasoning II
Professor: Craig DeLancey
Office: Marano 212A
Email: craig.delancey@oswego.edu

Current Assignments
27 February
Do problem 3 of chapter 20.

Some hints.

This is really 3 problems. After some reflection, I think this is the easiest order to do them in:
  1. derive (m3) in system S5
  2. derive (m4) in system S5
  3. derive (m5) in the combination of systems M and Brouwer.
1 and 2 will prove that from S5 you get Brouwer & M combined; and 3 will prove that from Brouwer & M combined you can get S5.

For 1: note that we've proved lots of times that (P → <>P). But (m5) is (<>P → []<>P). What does the chain rule give you?

For 2: this one is hard. To do it, why not fill in the blanks of this proof. I can't draw fitch bars, so we need to be careful about dependencies.
  1. (<>[]P → []P) this is the S5 dual, we prove a few times; you prove it again here
  2. [](<>[]P → []P) necessitation on the theorem on line 1
  3. ([](<>[]P → []P) → ([]<>[]P → [][]P) axiom (m2)
  4. ([]<>[]P → [][]P) modus ponens 2, 3

      Now note, we have proven that in S5 we have as a theorem axiom (m3) -- that is, we have (P → []<>P). An instance of this theorem is ([]P → []<>[]P). Chain rule!

      For 3: OK, this one is less hard, but still challenging. Some tricks will be required! Consider this. The following is a theorem we can prove using (m4): (<><>P --> <>P). I think we proved that but prove it again. Since it's a theorem, you have by necessitation that [](<><>P --> <>P) (for justification you can write, "theorem, necessitation"). Now, consider what you'll get using axiom (m2) if that is the antecedent of an instance of (m2) (remember that the antecedent is the 'if' part of a conditional--and (m2) is a conditional). Finally, note that this is an instance of (m3): (<>P-->[]<><>P). Think about chain rule.

Tentative assignments
2 March
Read chapter 1 of Mendelson.
4 March
We'll have some homework from Mendelson, using axioms.
9 March
We'll have some homework using mathematical induction.
23 March