Quiz Solutions

1. Yes it can: A! --> HA! (Rule 3) --> AHA!? (Rule 2) --> AHA! (Rule 4)

2a. 2 + 3 = 5 :

2 + S(2) = S(2 + 2) = S(2 + S(1)) = S(S(2 + 1)) = S(S(2 + S(0))) = S(S(S(2+0)))) = S(S(S(2))) = 5

2b. 1 x (2 + 3) = 5:

1 x (2 + S(2)) = 1 x (S(2+2)) = 1 x (S(2 + S(1)) = 1 x (S(S(2+1))) = 1 x (S(S(2 + S(0)))) = 1 x (S(S(S(2+0))))

= 1 x (S(S(S(2)))) = 1 x 5 = 5 x 1 = 5 x S(0) = 5 + (5 x 0) = 5 + 0 = 5

2c. 1 + (2 x 2) = 5:

1 + (2 x S(1)) = 1 + (2 + (2 x 1)) = 1 + (2 + (2 + S(0))) = 1 + (2 + (2 + (2 x 0))) = 1 + (2 + (2 + 0)) = 1 + (2 + 2) = 1 + (2 + S(1)) = 1 + S(2 + 1) = 1 + S(2 + S(0)) = 1 + S(S(2 + 0)) = 1 + S(S(2)) = 1 + 4 = 1 + S(3)

= S(1 + 3) = S(1 + S(2)) = S(S(1 + 2)) = S(S(1 + S(1))) = S(S(S(1 + 1))) = S(S(S(1 + S(0)))) = S(S(S(S(1 + 0))))

= S(S(S(S(1)))) = 5

3. An urelement is something that can be an element of a set but is itself not a set. The allowance of urelements in the New Foundations axiomatic set theory makes this system relatively consistent to Peano Arithmetic. Urelements also fit into the typing theory we saw in Principia Mathematica, acting as the lowest order type.