# How do we know that $c(n,a,G) = h^{\ast}(n)$, in the proof that if a heuristic is consistent then it is admissible?

I found a proof that if a heuristic $$h$$ is consistent, then it is admissible, but I'm confused by one of the steps in the proof.

The proof is by induction on $$k$$, the number of actions from a node $$n$$ to the goal node $$G$$.

Base step: $$k = 1$$, i.e., node $$n$$ is one step from $$G$$. Since the heuristic function $$h$$ is consistent,

$$h(n) \leq c(n, a, G) + h(G)$$

Since $$h(G) = 0$$,

$$h(n) \leq c(n, a, G) = h^{\ast}(n)$$

Therefore, $$h$$ is admissible.

Induction step: Assume: If $$h(n′)$$ is consistent, then $$h(n′)$$ is admissible, for all nodes $$n′$$ that are $$k$$ steps from $$G$$.

$$h(n) \leq c(n, a, n') + h(n')$$

Since $$n′$$ is $$k$$ steps from $$G$$ and $$h$$ is admissible for node $$n′$$ that is $$k$$ steps from $$G$$,

$$h(n') \leq h^{\ast}(n')$$ $$\implies c(n,a,n') + h(n') \leq c(n, a, n') + h^{\ast}(n')$$ $$\implies h(n) \leq c(n, a, n') + h^{\ast}(n') = h^{\ast}(n)$$

hence, $$h$$ us admissible for node n.

I'm specifically confused about these lines:

$$h(n) \leq c(n, a, G) = h^{\ast}(n)$$

and

$$h(n) \leq c(n, a, n') + h^{\ast}(n') = h^{\ast}(n)$$

How do we know that $$c(n,a,G) = h^{\ast}(n)$$? Wouldn't it depend on the action $$a$$ chosen?

Also, how do we know: $$c(n, a, n') + h^{\ast}(n') = h^{\ast}(n)$$?

$$c(n, a, G) = h^*(n)$$ is true because $$h^*$$ is the optimal heuristic, i.e. it gives you the minimal/optimal cost from $$n$$ to the goal $$G$$, which, in that base case, is exactly $$c(n, a, G)$$, i.e. the edge distance from $$n$$ to $$G$$.
Now, I understand the confusion arises from the parameter $$a$$, the action.
Here, $$a$$ is arbitrary, but, in this case, there's only one action that should lead to $$G$$, which I suppose is a reasonable assumption, otherwise you would have multiple arrows from $$n$$ to $$G$$ (which are neighbours), which I've never seen.
I don't even know why they use this parameter $$a$$. I think you can safely ignore it, unless you want to consider cases where there are more than one edge between two neighbours $$n$$ and $$n'$$, which is useless because you can always consider the cheapest edge.