General Computer Science 320201 GenCS I & II Lecture ... - Kwarc

+(〈n, o〉) = n (base) and +(〈m, s(n)〉) = s(+(〈m, n〉)) (step)
Theorem 107 + ⊆ (N1 × N1) × N1 is a total function.
We have to show that for all 〈n, m〉 ∈ (N1 × N1) there is exactly one l ∈ N1 with 〈〈n, m〉, l〉 ∈
+.
We will use functional notation for simplicity
Addition is a total Function
c○: Michael Kohlhase 75
Lemma 108 For all 〈n, m〉 ∈ (N1 × N1) there is exactly one l ∈ N1 with +(〈n, m〉) = l.
Proof: by induction on m. (what else)
P.1 we have two cases
P.1.1 base case (m = o):
P.1.1.1 choose l := n, so we have +(〈n, o〉) = n = l.
P.1.1.2 For any l ′ = +(〈n, o〉), we have l ′ = n = l.
P.1.2 step case (m = s(k)):
P.1.2.1 assume that there is a unique r = +(〈n, k〉), choose l := s(r), so we have
+(〈n, s(k)〉) = s(+(〈n, k〉)) = s(r).
P.1.2.2 Again, for any l ′ = +(〈n, s(k)〉) we have l ′ = l.
Corollary 109 +: N1 × N1 → N1 is a total function.
c○: Michael Kohlhase 76
The main thing to note in the proof above is that we only needed the Peano Axioms to prove
function-hood of addition. We used the induction axiom (P5) to be able to prove something about
“all unary natural numbers”. This axiom also gave us the two cases to look at. We have used the
distinctness axioms (P3 and P4) to see that only one of the defining equations applies, which in
the end guaranteed uniqueness of function values.
Reflection: How could we do this?
we have two constructors for N1: the base element o ∈ N1 and the successor function
s: N1 → N1
Observation: Defining Equations for +: +(〈n, o〉) = n (base) and +(〈m, s(n)〉) =
s(+(〈m, n〉)) (step)
the equations cover all cases: n is arbitrary, m = o and m = s(k)
(otherwise we could have not proven existence)
but not more (no contradictions)
using the induction axiom in the proof of unique existence.
Example 110 Defining equations δ(o) = o and δ(s(n)) = s(s(δ(n)))
Example 111 Defining equations µ(l, o) = o and µ(l, s(r)) = +(〈µ(l, r), l〉)
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