When I was solving a simple cycle problem, we assumed a function's symmetry about $x=a$ and $x=b$ and I tried to solve the function's cycle.
It seems very simple, and indeed it is. It's my process going about solving it. Because the function is symmetrical about $x=a$ and $x=b$.
So we have:
$$f(x-a)=f(x+a)\quad f(x-b)=f(x+b) \tag1$$
We let $x-a=t$, and then we can get:
$$f(t)=f(t+2a)\quad f(t)=f(t+2b)\tag2$$
Using $x$ to replace $t$, we get:
$$f(x)=f(x+2a)\quad f(x)=f(x+2b)\tag3$$
So we can let:
$$f(x+2a)=f(x+2b)\tag4$$
To get $f(x)=f(x+2(b-a))$. My question is why is formula $(4)$ valid? Does function have transitivity?
If we have $f(x),g(x),h(x)$,and $f(x)=g(x), g(x)=h(x)$, can we get $f(x)=g(x)$?