Frobenius number of $a,b,c$ where c is Frobenius of $a,b$.

Let $$g(a_1,...,a_n)$$ be Frobenius number of $$a_1,....a_n$$ where $$gcd(a_1,....a_n) = 1$$ and $$\Gamma(a_1,...,a_n) = \{x_1a_1+...+x_na_n : x_i \geq 0 \forall i\}$$. Then Frobenius number,$$g(a_1,...,a_n)$$,is the largest number such that $$g(a_1,...,a_n) \notin \Gamma(a_1,...,a_n)$$

According to A. Tripathi's paper on Frobenius on three variables,[Lemma 1] He shows that

if $$a,b$$ are coprime such that $$a and $$c = g(a,b)$$

$$g(a,b,c) = c - a$$

I try to understand and I have written the proof of my version as follow:

Note that, if $$n < c = g(a,b)$$, then $$n \in \Gamma(a,b,c)$$ if and only if $$n \in \Gamma(a,b).$$

Since $$(c-a) < c$$ and $$(c-a) \notin \Gamma(a,b)$$, $$(c-a) \notin \Gamma(a,b,c)$$. It leave to show that $$c-a$$ is the largest integer not in $$\Gamma(a,b,c)$$

Ok, there is a Theorem that $$g(a,b,c) = bx_0 + cy_0 - a$$ for some $$x_0,y_0\geq 0$$. We will show that $$x_0 = 0$$ and $$y_0 = 1$$.

if $$x_0,y_0\geq 1$$ $$bx_0 + cy_0 - a > (b-a) + c > c = g(a,b),$$

hence $$bx_0 + cy_0 - a \in \Gamma(a,b) \subset \Gamma(a,b,c)$$ contradiction. Since $$g(a,b,c) \notin \Gamma(a,b,c)$$

if $$y_0 \geq 2$$ and $$x_0 = 0$$, $$cy_0 - a > c + (c - a) > c = g(a,b),$$

hence $$cy_0 - a \in \Gamma(a,b) \subset \Gamma(a,b,c)$$

It leaves to show where $$x_0 \geq 1$$ and $$y_0 = 0$$ which I have been stuck for a while now for completing the proof.

Actually, A.Tripathi's proof uses another theorem (Theorem 1 in his paper) to claim but I do not get his argument also if someone could give me some explanations I'd be glad for your help.

The last case require the face that $$g(a,b) = ab - a - b$$
suppose that $$g(a,b,c) = bx_0 + cy_0 - a \geq c - a$$ where $$x_0\geq 1$$ and $$y_0 = 0$$ then $$g(a,b,c) = bx_0 - a \geq c - a$$. So $$bx_0 \geq c = g(a,b) =ab-a-b$$
hence $$x_0 \geq a -\frac{a}{b} - 1$$. Since $$x_0$$ must be integer greater that $$a -\frac{a}{b} - 1$$, $$x_0 \geq a-1$$
And then, $$bx_0 \geq ba - b = c + a$$.
So, $$bx_0 -a \geq c$$.
But we assume that $$g(a,b,c) = bx_0 - a$$ and by above give that $$g(a,b,c) \geq c$$ but If $$g(a,b,c) = c$$ then $$g(a,b,c)\in \Gamma(a,b,c)$$ and if $$g(a,b,c) > c$$ $$g(a,b,c)\in \Gamma(a,b) \subset \Gamma(a,b,c)$$ contradiction.