Proving a recurrence relation by induction $f(L) = f(L-1) +1 = L-1.$

Question

I'm trying to prove inductively that the recurrence relation :

$f(L) = f(L-1)+1$ is equivalent to $g(L) = L-1$

with some following Base cases: $f(0) = 0$, $f(1) = 0$, $f(2) = 1.$ for the recurrence relation.

So far: $ L = 1$

$f(1) = 0$ because of the base case, and $g(1) = 1 - 1 = 0 $ So it is TRUE at $L = 1$

$L = k$

$f(k) = f(k-1) + 1$ and $g(k) = k-1$

Answer 1

If $f(k) = k-1$, then $f(k+1) =f(k)+1 =(k-1)+1 =k =(k+1)-1 $ so it is true for $k+1$.

Since it is true for $k=1$, it is true for all $k \ge 1$.