Let $\Phi (x)$ be the cumulative distribution function of the standard normal distribution. Given $x_0$, $x_1 = \Phi(x_0-x_1)$.If $x_n$ is given, $x_{n+1} = \Phi(x_{n}-x_{n+1})$(By drawing a graph, $\lim_{n \to \infty}{x_n} = 1/2$, for all $x_0 \in \mathbb{R}$).
My question is how to show that $\lim_{n \to \infty}{x_n}$ exists and is unique in the sense that it doesn't depend on the value of $x_0$ by contraction mapping principle.
The problem is that I don't know how to deal with the functions in the form of $F(x,y) = 0$, which can't be converted explicitly to the form $y = f(x)$. A general method is especially welcome.
Assume that the function $F(x,y)$ is continuously differentiable and that the partial derivative $F_y $ never turns to $0$: $$F\in C^1(\mathbb R^2),\qquad F_y(x,y)\ne 0 \quad \forall (x,y)\in\mathbb R^2 \tag1 $$ Then $F_y$ has constant sign, which implies that for any fixed $x$ the equation $F(x,y)=0$ has at most one solution. To ensure that there is a solution, we should check that $$\forall x\in\mathbb R \ \text{ the function } y\mapsto F(x,y) \ \text{takes positive and negative values} \tag2 $$ If (1) and (2) hold, then the equation $F(x,y)=0$ defines a function $y=f(x)$ with $f:\mathbb R\to\mathbb R$.
The 1-dimensional case of Implicit Function theorem (unfortunately not spelled out on Wikipedia explicitly) says that $f$ is continuously differentiable and its derivative can be computed as $$f'(x)=-\frac{F_x}{F_y}$$ Therefore, to apply the contraction mapping theorem to $f$, we remains to check that $$|F_x|\le \lambda |F_y| \quad \text{ everywhere } \tag3$$
All of the above is easy to carry out for the function $F(x,y)=y-\Phi(x-y)$ in your example. The value of $\lambda$ turns out to be something mildly entertaining, like $1/(1+\sqrt{2\pi})$.