A function $g:[a,b]→\mathbb{R}$ is increasing if $∀x_1{<}x_2$ in $[a,b]$, $g(x_1)≤g(x_2).\quad$ Similar is the definition of a decreasing function.
So, if a function $f:[a,b]→\mathbb{R}$ satisfies property $p:\quad f$ is neither increasing nor decreasing, then:

My conclusion ($c_1$):
$∃x_1{<}x_2$ and $x_3{<}x_4$ in $[a,b]$ which satisfy $f(x_1)<f(x_2)$ and f$(x_3)>f(x_4)$ (or $f(x_1)>f(x_2)$ and $f(x_3)<f(x_4)$). That is, $f$ 'increases' for some $x_1$,$x_2$ and 'decreases' for some $x_3,x_4$ (or vice versa).

My textbook's and instructor's conclusion ($c_2$):
$∃x_1{<}x_2{<}x_3$ in $[a,b]$ s.t.$f(x_1)<f(x_2)$ and f$(x_2)>f(x_3)$ (or viceversa).
My questions are:

I think that $c_1$ can be logically concluded from $p$, but not $c_2.$
Which conclusion can be derived immediately from $p$ by logic?

Are the $c_1$ and $c_2$ identical? If yes, how can we prove it?
There are a few theorems and proofs that can be easily proved from $c_2$, not $c_1$. But I think that the proofs which use $c_2$ are not complete, but neither I can prove those theorems using $c_1$, nor prove $c_2$ from $c_1$, so I need some help.
Best Answer
I suspect you're missing the "(or vice versa)" clause at the end of the conclusion $(c_2)$. Without that clause, $(c_2)$ certainly wouldn't follow from property $(p)$.
However, with that clause, since $\ f\ $ is not decreasing over $\ [a,b]\ $, there must exist $\ y_1,y_2\in[a,b]\ $ such that $\ y_1<y_2\ $ and $\ f\big(y_1\big)<$$\,f\big(y_2\big)\ $, and since $\ f\ $ is not increasing, one of the following four conditions must hold:
In the first case, the stated condition of $(c_2)$ holds with $ x_i=y_i\ $, and in the second, it holds with $\ x_1=y_1,$$\ x_2=y_3\ $ and $\ x_3=y_2\ $. In the third case the viceversa condition holds with $\ x_1=y_1, $$\ x_2=y_3\ $ and $\ x_3=y_2\ $. That is, $\ f\big(x_1\big)>$$\, f\big(x_2\big)\ $ and $\ f\big(x_2\big)<$$\, f\big(x_3\big)\ $ with $\ x_1<x_2<x_3\ $. In the fourth case, the same viceversa condition holds with $\ x_1=y_4,$$\ x_2=y_3\ $ and $\ x_3=y_2\ $.