# Daily Challenge #168 - [Code golf] f (f (f b)) = f b

The kata is inspired by a Stack Overflow question.

It is easy to prove that f(f(f b)) = f b for all functions f : bool -> bool. But can you do it in less than 92 characters?

Lemma lemma : forall (f : bool -> bool) (b : bool), f (f (f b)) = f b.

And the size of your solution (including all declarations) should be 91 characters or less.

This challenge comes from monadius on CodeWars. Thank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!

Want to propose a challenge idea for a future post? Email yo+challenge@dev.to with your suggestions!

### Discussion

joomy

Here's my solution in Coq:

Lemma lemma : forall (f : bool -> bool) (b : bool), f (f (f b)) = f b.
intros;case b eqn:G,(f b) eqn:H,(f (f b)) eqn:I;rewrite G,H,I in*;intuition. Qed.


The second line has 80 characters.

Daan Wilmer

Not a code golfer, but pretty compact with writing the way I learned in uni:

f(x)=c → f(f(f(x)))=c=f(x)
f(x)=x → f(f(f(x)))=x=f(x)
f(x)=¬x → f(f(f(x)))=¬¬¬x=¬x=f(x)


Above code is 82 characters, excluding whitespace.

EDIT: now in LaTeX

\begin{aligned} f(x) = c &\rightarrow f(f(f(x))) = c = f(x) \\ f(x) = x &\rightarrow f(f(f(x))) = x = f(x) \\ f(x) = \neg x &\rightarrow f(f(f(x))) = \neg \neg \neg x = \neg x = f(x) \end{aligned}
José A. Alonso

Another proof in Isabelle/HOL

lemma
fixes f :: "bool ⇒ bool"
shows "f (f (f b)) = f b"
by (cases b; cases "f True"; cases "f False"; simp)

José A. Alonso

In Isabelle/HOL

lemma "(f :: bool ⇒ bool) (f (f b)) = f b"
by smt

joomy

You have a typo in the lemma statement, f is applied twice instead of three times.

José A. Alonso

In Isabelle/HOL equivalent to the above but with a different notation

lemma "∀(f :: bool ⇒ bool). f (f (f b)) = f b"
by smt

lemma