I thought it might be helpful to do some exercises! We'll be sweatin' to the oldies. Or something. Everyone get out your leotards and legwarmers!

Most of these are pulled from the notes and the book. I'll post solutions after the recitation (so you stinkers in the back withthe laptops can't just read off the answers to me. :P )

Oh and ⇒ is supposed to show up as the implication symbol (->). Hopefully your browser supports this?

Prove:




P ⇒ S, given (P ⇒ (R ⇒ S)), (Q ⇒ R), and (P ⇒ Q).






(P ⇒ Q) ⇒ ((Q ⇒ R) ⇒ (P ⇒ R))






Solution:
                 (P ⇒ Q)z    Px
(Q ⇒ R)y                Q 
               R    x
            P ⇒ R    y
        (Q ⇒ R) ⇒ (P ⇒ R)  z
  (P ⇒ Q) ⇒ ((Q ⇒ R) ⇒ (P ⇒ R))
    







(M ⇒ O) ⇒ (N ⇒ O), assuming you're given a proof of M.











Q ⇒ (P ⇒ P)













P ⇒ (Q ⇒ P)












P ⇒ (Q ⇒ R)) ⇒ ((P ⇒ Q) ⇒ (P ⇒ R))
















P ⇒ ((P ⇒ (Q ⇒ R)) ⇒ ((P ⇒ Q) ⇒ (P ⇒R)))











P ⇒ ((P ⇒ Q) ⇒ Q)
Now, say that P is the proposition (R ⇒ R).
Prove that R ⇒ R,
then use elimination to prove
((R ⇒ R) ⇒ Q) ⇒ Q





Show that this proof is quite ridiculously redundant, and you can actually prove ((R ⇒ R) ⇒ Q) ⇒ Q) in four steps, instead of six.
Allons-y! Back to CIS160 Page