Last Update: 12 November 2010
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…Previous lecture
that's just an easier way of typing:

Suppose P(k), & try to show P(k+1).
Note: The antecedent "P(k)" is called the 

Thm (General Distributive Law)
proof:
The "property" P that we're interested in proving that it holds for all natural numbers is:
Base Case (first (failed) attempt):
Base Case (second (slightly more successful) attempt):
That's trivially true! That was easy!
There's only one problem: It won't be of any help to us :
So, let's restate the theorem one more time, as:
(This is the only interesting case anyway.)
So now we're ready for the…
…Inductive Case:
i.e., show ∀k[ (a(b_{1}+…+b_{k}) = ab_{1}+…+ab_{k}) → (a(b_{1}+…+b_{k+1}) = ab_{1}+…+ab_{k+1}) ]
& show the consequent: a(b_{1}+…+b_{k+1}) = ab_{1}+…+ab_{k+1}:
Let's compute a(b_{1}+…+b_{k+1}):
= a( (b_{1}+…+b_{k}) + b_{k+1})
= a( B + b_{k+1})
= aB + ab_{k+1}, by the base case!
= a(b_{1}+…+b_{k}) + ab_{k+1}
= (ab_{1}+…+ab_{k}) + ab_{k+1}
QED
Thm (the cardinality of the power set of S = 2 raised to the cardinality of S)
proof:
Base Case:
[This is as far as we got in lecture; now for the rest of the proof:]
Inductive Case:
Describing this in FOL, we have:
∃S′′[S′=S′′∪{e} ∧ S′′=k]
∴ ℘(S′′)=2^{k}, by the inductive hypothesis
But to each subset of S′′, there correspond two subsets of S′:
i.e., in (sort of) FOL:
(∀ subset S^{*} ⊆ S′′)(∃ 2 subsets of S′=S′′∪{e}), namely: S^{*} itself & S^{*}∪{e}.
∴ ℘(S′) = ℘(S′&prime)*2 (i.e.,
doubled)
= 2^{k}*2 = 2^{k+1}
QED