Fully invariant implies verbal in reduced free group

Verbal statement
In a reduced free group, every fully invariant subgroup is verbal.

Statement with symbols
Suppose $$G$$ is a reduced free group, i.e., $$G$$ is the quotient of a free group by a verbal subgroup. Then, in $$G$$, any fully invariant subgroup is verbal.

Converse
The converse statement is true in any group: a verbal subgroup is always a fully invariant subgroup.

Facts used
In any variety of algebras, if $$S$$ is a freely generating set for an algebra $$F$$, and $$A$$ is any other algebra in that variety, any set-theoretic map from $$S$$ to $$A$$ extends uniquely to a homomorphism from $$F$$ to $$A$$.

In particular, if $$G$$ is reduced free, then any set-theoretic map from a (reduced)-freely generating set of $$G$$ to $$G$$, extends uniquely to an endomorphism of $$G$$.

Proof
Given: A reduced free group $$G$$, a fully invariant subgroup $$H$$ of $$G$$

To prove: There exists a set $$W$$ of words, such that $$H$$ is precisely the set of elements of $$G$$ expressible using words from $$W$$, by substituting elements from $$G$$

Proof: Since $$G$$ is reduced free, we can take a set of generators $$x_i, i \in I$$ for $$G$$ coming from a freely generating set. Define $$W$$ as the set of all words $$w$$ for which there exists $$h \in H$$ where $$h = w(x_{j_1}, x_{j_2}, \dots, x_{j_r})$$ for $$j_l \in I$$. We claim that $$W$$ is as required. For this, we observe two things:

One direction: trivial
Clearly, every element of $$H$$ is expressible using a word in $$W$$, substituting elements of $$G$$

Other direction
Consider $$w(g_1,g_2,\dots,g_n)$$ for some $$w \in W$$ and $$g_i \in G$$. We want to show that $$w(g_1,g_2,\dots, g_n) \in H$$.

We know that since $$w \in W$$, there exist $$x_{j_1}, x_{j_2}, \dots, x_{j_n}$$ such that $$w(x_{j_1},x_{j_2},\dots,x_{j_n}) \in H$$. Consider the map $$x_{j_i} \mapsto g_i$$ and extend this to an arbitrary map from the set $$\{ x_i \}$$ to the group $$G$$. By the universal property, this set-theoretic map extends to an endomorphism $$\varphi$$ of $$G$$, such that:

$$\varphi(x_{j_i}) = g_i$$

Since endomorphisms preserve words, we get:

$$\varphi(w(x_{j_1},x_{j_2},\dots,x_{j_n})) = w(g_1,g_2,\dots,g_n)$$

Now, $$w(x_{j_1},x_{j_2},\dots,x_{j_n}) \in H$$ and $$H$$ is a fully invariant subgroup, so $$\varphi(w(x_{j_1},x_{j_2},\dots,x_{j_n})) \in H$$, hence $$w(g_1,g_2,\dots,g_n) \in H$$, and we are done.