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

step condition go through: we we can use the stronger inductive hypothesis in the proof of step
case, which is simpler.
Often the key idea in an induction proof is to find a suitable strengthening of the assertion to
get the step case to go through.
What can we do with unary natural numbers?
So far not much (let’s introduce some operations)
Definition 38 (the addition “function”) We “define” the addition operation ⊕ procedurally
(by an algorithm)
adding zero to a number does not change it.
written as an equation: n ⊕ o = n
adding m to the successor of n yields the successor of m ⊕ n.
written as an equation: m ⊕ s(n) = s(m ⊕ n)
Questions: to understand this definition, we have to know
Is this “definition” well-formed? (does it characterize a mathematical object?)
May we define “functions” by algorithms? (what is a function anyways?)
c○: Michael Kohlhase 38
Addition on unary natural numbers is associative
Theorem 39 For all unary natural numbers n, m, and l, we have n⊕(m ⊕ l) = (n ⊕ m)⊕l.
Proof: we prove this by induction on l
P.1 The property of l is that n ⊕ (m ⊕ l) = (n ⊕ m) ⊕ l holds.
P.2 We have to consider two cases base case:
P.2.1.1 n ⊕ (m ⊕ o) = n ⊕ m = (n ⊕ m) ⊕ o
P.2.2 step case:
P.2.2.1 given arbitrary l, assume n⊕(m ⊕ l) = (n ⊕ m)⊕l, show n⊕(m ⊕ s(l)) = (n ⊕ m)⊕
s(l).
P.2.2.2 We have n ⊕ (m ⊕ s(l)) = n ⊕ s(m ⊕ l) = s(n ⊕ (m ⊕ l))
P.2.2.3 By inductive hypothesis s((n ⊕ m) ⊕ l) = (n ⊕ m) ⊕ s(l)
c○: Michael Kohlhase 39
More Operations on Unary Natural Numbers
Definition 40 The unary multiplication operation can be defined by the equations n⊙o = o
and n ⊙ s(m) = n ⊕ n ⊙ m.
Definition 41 The unary exponentiation operation can be defined by the equations
exp(n, o) = s(o) and exp(n, s(m)) = n ⊙ exp(n, m).
Definition 42 The unary summation operation can be defined by the equations o i=o ni =
o and s(m) i=o ni = ns(m) ⊕ m i=o ni.
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