P310 Valid Reasoning II
Professor: Craig DeLancey
Office: Marano 212A
Do problem 3 of chapter 20.
This is really 3 problems. After some reflection, I think this
is the easiest order to do them in:
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.
- derive (m3) in system S5
- derive (m4) in system S5
- derive (m5) in the combination of systems M and Brouwer.
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.
- (<>P → P) this is the S5 dual, we prove a few times; you prove it again here
- (<>P → P) necessitation on the theorem on line 1
- ((<>P → P) → (<>P → P) axiom (m2)
- (<>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
Read chapter 1 of Mendelson.
We'll have some homework from Mendelson, using axioms.
We'll have some homework using mathematical induction.