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

thepracticaldev profile image dev.to staff ・1 min read

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?

More specifically, your task is to prove the following lemma:

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

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cattheory profile image
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.

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dwilmer profile image
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

f(x)=cf(f(f(x)))=c=f(x)f(x)=xf(f(f(x)))=x=f(x)f(x)=¬xf(f(f(x)))=¬¬¬x=¬x=f(x) \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}
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jose_a_alonso profile image
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)
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jose_a_alonso profile image
José A. Alonso

In Isabelle/HOL

lemma "(f :: bool ⇒ bool) (f (f b)) = f b"
  by smt
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cattheory profile image
joomy

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

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jose_a_alonso profile image
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
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jose_a_alonso profile image
José A. Alonso

And with another notation

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