100 in Golden Mean Number System is Equivalent to 011

Theorem
Consider the golden mean number system.

Let the restrictions:
 * $(1): \quad$ No two adjacent $1$s may appear in the representation
 * $(2): \quad$ The representation may not end with the infinite sequence $\cdotp \cdots 010101 \ldots$

be lifted.

Let $p$ and $q$ be arbitrary strings in $\left\{ {0, 1}\right\}$.

Let $x \in \R_{\ge 0}$ have a representation which includes the string $100$, say:
 * $x = p100q$

Then $x \in \R_{\ge 0}$ also has the representation:
 * $x = p011q$

Similarly, let $x \in \R_{\ge 0}$ have a representation which includes the string $011$, say:
 * $x = p011q$

Then $x \in \R_{\ge 0}$ also has the representation:
 * $x = p100q$

That is, any instance of $100$ appearing in a representation of a non-negative real number $x$ can be replaced with $011$ without changing $x$, and vice versa.

Note that the instance of $100$ or $011$ may also include a radix point; the instance of $011$ or $100$ that replaces it will include the radix point in the same location.

Proof
Let $100$ appear anywhere within $x$.

Then:
 * $x = \phi^r + \displaystyle \sum_{c \mathop \in C} \phi^c$

where:
 * $C \subset \Z$
 * $r \in \Z$
 * $r \notin C, r - 1 \notin C, r - 2 \notin C$

That is, the instance of $100$ corresponds to the indices $r, r - 1, r - 2$.

From Power of Golden Mean as Sum of Smaller Powers:
 * $\phi^r = \phi^{r - 1} + \phi^{r - 2}$

and so:
 * $x = \phi^{r - 1} + \phi^{r - 2} + \displaystyle \sum_{c \mathop \in C} \phi^c$

That is, the indices instance of $100$ corresponds to the indices $r, r - 1, r - 2$ now correspond to the string $011$.