Constant Function is Primitive Recursive

Theorem
The constant function $$f_c: \N \to \N$$, defined as:
 * $$f_c \left({n}\right) = c$$ where $$c \in \N$$

is primitive recursive‎.

Generalization
The constant function of $$k$$ variables: $$f^k_c: \N^k \to \N$$, defined as:
 * $$f^k_c \left({n_1, n_2, \ldots, n_k}\right) = c$$ where $$c \in \N$$

is primitive recursive‎.

Proof by induction
First we note that $$f_0: \N \to \N$$ is the zero function, which is a basic primitive recursive function.

Base Case
Next we show that $$f_1: \N \to \N$$ is primitive recursive‎, as follows.

The successor function $$\operatorname{succ}: \N \to \N$$, defined as:
 * $$\forall n \in \N: \operatorname{succ} \left({n}\right) = n + 1$$

is a basic primitive recursive function.

Since $$\operatorname{succ} \left({0}\right) = 1$$, we have that:
 * $$f_1 \left({n}\right) = \operatorname{succ} \left({\operatorname{zero} \left({n}\right)}\right)$$.

Thus $$f_1$$ is obtained from the basic primitive recursive functions $$\operatorname{succ}$$ and $$\operatorname{zero}$$ by substitution.

So $$f_1: \N \to \N$$ is primitive recursive‎.

This is our base case.

Induction Hypothesis
This is our induction hypothesis:
 * $$f_k: \N \to \N$$ is primitive recursive‎ for some $$k \in \N$$.

Then we need to show:
 * $$f_{k+1}: \N \to \N$$ is primitive recursive‎.

Induction Step
This is our induction step:
 * $$f_{k+1} \left({n}\right) = k+1 = \operatorname{succ} \left({k}\right) = \operatorname{succ} \left({f_k \left({n}\right)}\right)$$.

Now $$f_k \left({n}\right)$$ is primitive recursive‎ from our induction hypothesis.

Thus $$f_{k+1} \left({n}\right)$$ is obtained from the basic primitive recursive function $$\operatorname{succ}$$ and $$f_k \left({n}\right)$$ by substitution.

The result follows by the Principle of Mathematical Induction.

Therefore $$\forall c \in \N: f_c \left({n}\right) = c$$ where $$c \in \N$$ is primitive recursive‎.

Proof of Generalization
For $$k \ge 1$$, let $$f^k_c$$ be the constant function of $$k$$ variables with value $$c$$.

We already know from the main proof that $$f^1_c$$ is primitive recursive‎.

Now:
 * $$f^k_c \left({n_1, n_2, \ldots, n_k}\right) = f^1_c \left({n_1}\right) = f^1_c \left({\operatorname{pr}^k_1 \left({n_1, n_2, \ldots, n_k}\right)}\right)$$

where $$\operatorname{pr}^k_1$$ is a projection function which is basic primitive recursive.

So $$f^k_c$$ is obtained from the primitive recursive‎ function $$f^1_c$$ and the basic primitive recursive function $$\operatorname{pr}^k_1$$ by substitution.

Hence by definition, $$f^k_c$$ is primitive recursive‎.