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# [2018-Nov] Triple Rev

Is there a function rev of type "nat list => nat list" that applied thrice on a list of nats returns the original list again.

i.e. find a function rev and prove the following lemma:

rev (rev (rev xs)) = xs

Making sure you do not use the identity function, if your input is longer than 1.

length xs >= 2 ==> rev xs ~= xs

## Resources

### Definitions File

```theory Defs
imports Main
begin

end
```

### Template File

```theory Submission
imports Defs
begin

lemma showmewhatyougot: "\<exists>f::(nat list \<Rightarrow> nat list). (\<forall> xs. (length xs \<ge> 2 \<longrightarrow> f xs \<noteq> xs))
\<and> (\<forall> xs. (f (f (f xs))) = xs)"
sorry

end
```

### Check File

```theory Check
imports Submission
begin

lemma  "\<exists>f::(nat list \<Rightarrow> nat list). (\<forall> xs. (length xs \<ge> 2 \<longrightarrow> f xs \<noteq> xs))
\<and> (\<forall> xs. (f (f (f xs))) = xs)"
by (rule Submission.showmewhatyougot)

end
```

### Definitions File

```theory Defs
imports Main
begin

end
```

### Template File

```theory Submission
imports Defs
begin

lemma showmewhatyougot: "\<exists>f::(nat list \<Rightarrow> nat list). (\<forall> xs. (length xs \<ge> 2 \<longrightarrow> f xs \<noteq> xs))
\<and> (\<forall> xs. (f (f (f xs))) = xs)"
sorry

end
```

### Check File

```theory Check
imports Submission
begin

lemma  "\<exists>f::(nat list \<Rightarrow> nat list). (\<forall> xs. (length xs \<ge> 2 \<longrightarrow> f xs \<noteq> xs))
\<and> (\<forall> xs. (f (f (f xs))) = xs)"
by (rule Submission.showmewhatyougot)

end
```

### Definitions File

```theory Defs
imports Main
begin

end
```

### Template File

```theory Submission
imports Defs
begin

lemma showmewhatyougot: "\<exists>f::(nat list \<Rightarrow> nat list). (\<forall> xs. (length xs \<ge> 2 \<longrightarrow> f xs \<noteq> xs))
\<and> (\<forall> xs. (f (f (f xs))) = xs)"
sorry

end
```

### Check File

```theory Check
imports Submission
begin

lemma  "\<exists>f::(nat list \<Rightarrow> nat list). (\<forall> xs. (length xs \<ge> 2 \<longrightarrow> f xs \<noteq> xs))
\<and> (\<forall> xs. (f (f (f xs))) = xs)"
by (rule Submission.showmewhatyougot)

end
```

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