Proving the composition of an even function and an even function is even using the given definitions

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There are multiple answers to this question, but they use a slightly different version of what it means to be an odd function. I wanted to know what I did wrong in my process of trying to arrive at a solution.

Defn: A function $f$ is odd if $f(x) = -f(-x)$ and a function is even if $f(x) = f(-x)$

Suppose $f$ is even and $g$ is even:

$$(f \circ g)(x) = f(g(x)) \\ = f(g(-x)) \ (\text{by definition of $g$ being even}) \\ = f(-g(-x)) \ (\text{by definition of $f$ being even}) \\ = f(-1 \cdot g(-x)) $$

by properties of scalars with functions:

$$ = -f(g(-x)) \\ = -(f \circ g)(-x) $$

I know that this is wrong, and I feel it must be in the 4th line that I did something wrong, what is it that is incorrect?

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The phrase "by properties of scalars with functions" and the appearance of this $1$ make things a bit confusing. But the main issue is that you have an extra minus sign. So instead, you would like to write $$ f(g(-x)) = f(-g(x)) = f(g(x)). $$ The first equality uses that $g$ is odd, and the second that $f$ is even.