--- /dev/null
+1. Find a fixed point `X` for the succ function:
+
+ H := \f.succ(ff)
+ X := HH
+ == (\f.succ(ff)) H
+ ~~> succ(HH)
+
+So `succ(HH) <~~> succ(succ(HH))`.
+
+2. Prove that `add 1 X <~~> X`:
+
+ add 1 X == (\mn.m succ n) 1 X
+ ~~> 1 succ X
+ == (\fz.fz) succ X
+ ~~> succ X
+
+What about `add 2 X`?
+
+ add 2 X == (\mn.m succ n) 2 X
+ ~~> 2 succ X
+ == (\fz.f(fz)) succ X
+ ~~> succ (succ X)
+
+So `add n X <~~> X` for all (finite) natural numbers `n`.
+
+
+