TITLE: Proof that $\sin(x(t))$ is not a linear and time-invariant system?
QUESTION [5 upvotes]: Suppose $\sin(x(t))$. I want to prove whether $\sin(x(t))$ is both linear and time-invariant or not. Is this proof flawed, or is it sound and correct?
Time Invariance:

Suppose $x_1(t)$ is a particular input signal to the system.
Now, suppose $x_2(t)$ is $x_1(t)$ but shifted by $T$ time units such that $x_2(t) = x_1(t - T)$.
Suppose $y_1(t)$ and $y_2(t)$ are the responses of the input signals of the system. 
To prove time-invariance, we must determine whether $y_1(t - T) = y_2(t)$ holds or not.

Therefore, we expand both sides:
$$LHS: y_1(t - T) = \sin(x_1(t - T))$$
$$RHS: y_2(t) = \sin(x_2(t))$$
$$sin(x_1(t - T))$$
Lemma 1: Therefore, $\sin(x(t))$ is a time-invariant system.

Linearity

Suppose $x_1(t)$ and $x_2(t)$ represent two distinct input signals, and $y_1(t)$ and $y_2(t)$ represent the two responses of the input signals into the system, respectively. Therefore...

$$y_1(t) = \sin(x_1(t))$$
$$y_2(t) = \sin(x_2(t))$$

To prove linearity, we must prove two properties: the scaling property and the additive property.

Scaling

To prove the scaling property, we must determine whether the following equation is true:

$$Ay(t) = \sin( Ax(t))$$
such that $A \in \mathbb{R}$.

After substitution...

$$A \sin(x(t)) \neq \sin(Ax(t))$$

Therefore, the system does not satisfy the scaling property.
Therefore, the system is not linear.
Therefore, the system is not both linear and time-invariant.

REPLY [3 votes]: It is correct. You could also find out about the nonliterary by looking at the Taylor series expansion of $\sin(x(t))$:
$$\sin(x(t))=x(t)-\frac{x^3(t)}{3!}+\frac{x^5(t)}{5!}+\cdots$$
the higher degrees of $x(t)$ are all nonlinear.