I need to formal proof that below algorithm return 1 for n = 1 and 0 in other cases.

```
function K( n: word): word;
begin
if (n < 2) then K := n
else K := K(n − 1) * K(n − 2);
end;
```

Anyone could help? Thank you

I need to formal proof that below algorithm return 1 for n = 1 and 0 in other cases.

```
function K( n: word): word;
begin
if (n < 2) then K := n
else K := K(n − 1) * K(n − 2);
end;
```

Anyone could help? Thank you

This can be proven by induction, but as previous posters have shown, it's tricky to get formally correct when referring to `K`

directly in the proof.

Here's my suggestion: Let *P*(*n*) be the property we want to show:

*P*(*n*) holds iff *K*(*n*) yields 1 for *n* = 1, and 0 for *n* ≠ 1.

Now we can clearly express what we want to show: Ɐ*n*.*P*(*n*)

**Base case:***n*≤ 2Trivial check by case analysis:

*P*(0) is ok, since*K*(0) = 0

*P*(1) is ok, since*K*(1) = 1**Induction hypothesis:***P*(*n*) holds for all 2 ≤*n*<*c*.**Inductive step:**Show that*P*(*c*) holds- By definition of
*K*we have*K*(*c*) =*K*(*c*-1) ×*K*(*c*-2) - By the induction hypothesis, we know that
*P*(*c*-1) and*P*(*c*-2) hold. - Since at most one of
*K*(*c*-1) and*K*(*c*-2) can be 1 (and the other must be 0) the product is 0. - Which means that
*P*(*c*) holds

- By definition of

Qed.

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