CSE 607
Natural Numbers and Lists Exam
Exam # 4 May 6, 2002
This is an open-book and open-notes examination. Do all problems in 80 minutes. Make sure you
justify your steps using the formal inference rules to earn full credit.
1. (10 points) Consider the following theory of natural numbers
A0. x = x
A1. (SUC(x : nat) =0)
A2. SUC(x : nat) = SUC(y : nat) x=y
A3. x:nat + 0 =x
A4. x:nat + SUC(y : nat) = SUC(x+y)
Prove 3≠2.
2. (10 points) Give tableau proof of the following
A0. x=x
A1. ¬(SUC(x : nat) = 0)
A2. SUC(x : nat) = SUC(y : nat) x = y
A3. x: nat + 0 = x
A4. x: nat + SUC(y : nat) = SUC (x+y)
A5. x : nat + 1 = SUC(x)
A6. x : nat + y : nat = y + x
A7. x : nat • 0 = 0
A8. x : nat • SUC(y : nat) = x•y + x
A9. x : nat • 1 =x
A10. 0 • x : nat =0
A11. 1 • x:nat = x
A12. SUC(x : nat) • y : nat = x • y + y
A13. (x : nat + y : nat) • z : nat = x•z + y • z
G1: ∀x y z [(x•y)•z = x • (y • z)]
3. (15 points) Consider the following tableau where APPEND is denoted by ++. You may
!"$#!% &%('*) #(+)", -
list, etc.
A0. x = x
A1. ¬([ ] = h::t)
A2. ((h1::t1) = (h2::t2)) (h1 = h2) ^ (t1 = t2)
A3. [ ] ++ l = l
A4. (h :: t) ++ l = h :: (t ++ l)
A5. l ++ [ ] = l
A6. l1 ++ ( l2 ++ l3) = (l1 ++ l2) ++ l3
A7. MAP f [ ] = [ ]
A8. MAP f (h :: t) = f(h) :: (MAP f t)
G1. ∀l1 l2 f . MAP f(l1 ++ l2) = MAP f l1 + MAP f l2
