$$f(0) = 1\\ h(0) = 1\\ g(0) = 0\\ f(n + 1) = 1 − h(n)\\ h(n + 1) = 1 − g(n + 1)\\ g(n + 1) = f(n) $$
Prove using induction that $∀n ∈ \Bbb N: f(n) + g(n) = 1$
what i've done so far:
Base Case: $n=0$ $$ f(0) + g(0) = 1\\ 1 + 0 = 1\\ 1 = 1 $$
Step Case: $$ f(n+1) + g(n+1) = 1\\ 1 - h(n) + f(n) = 1\\ 1 - h(0) + f(0) = 1\\ 1 - 1 + 1 = 1\\ 1 = 1 $$ Is this the right way to do it?
Once you have done the base case, let's look at the induction step. Assume that $f(k) + g(k) = 1$ for all $k\leq n$. Then we have \begin{align*} f(n+1) + g(n+1) &= 1 - h(n) + f(n)\\ &= 1 - (1- g(n)) + f(n)\\ &= g(n) + f(n) \\ &= 1 \end{align*} Because $f(n) + g(n) = 1$ by assumption.