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

Daily Challenge (273 Part Series)

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!

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Daily Challenge (273 Part Series)

### Discussion 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.

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}

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)


In Isabelle/HOL

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


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

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

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


And with another notation

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