Slide 14
Slide 14 text
14
For all sets , > | |. Bogus non-proof:
Proof by induction:
∷= ∀ sets where = . > .
Base case: 0 : = ∅.
= {∅}. = 1 > = 0.
Inductive case: ∀ ∈ ℕ. ⇒ + 1 .
for all sets where = + 1,
∃ where = , ∉ . = ∪
⇒ > ⇒ + 1 > + 1
Since () includes all elements of () and includes , this means
> ⇒ ( + 1).
Therefore, always holds, so we can conclude for all sets , > | |.