A bird L
is called a Lark
if, for any birds x
and y
, the following condition holds:
L(x)(y) => x(y(y));
Prove that if the forest contains a Lark L
and an Identity bird I
, it must also contain a Mockingbird M
.
L(x)
returns a Mockingbird! It evaluates to y => I(y(y))
, which (because the Identity bird doesn't modify its argument) is the same as y => y(y)
.