Logical conclusion from a function being neither increasing nor decreasing. 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 vice-versa).
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.
 A: 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:

*

*there exists $\ y_3>y_2\ $ such that $\ f\big(y_ 2\big)> f\big(y_ 3\big)\ $;

*there exists $\ y_3\in\big(y_1,y_2\big)\ $ such that $\ f\big(y_ 3\big)> f\big(y_ 2\big)\ $;

*there exists $\ y_3\in\big(y_1,y_2\big)\ $ such that $\ f\big(y_ 1\big)> f\big(y_3\big)\ $; or

*There exists $\ y_3\le y_1 $ and $\ y_4<y_3\ $ with $\ f\big(y_3\big)\le f\big(y_1\big)\ $ and $\ f\big(y_ 4\big)> f\big(y_3\big)\ $.

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,$$\ x_2=y_3\ $ and $\ x_3=y_2\ $. In the third case the vice-versa condition holds with $\ x_1=y_1, $$\ 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 vice-versa condition holds with $\ x_1=y_4,$$\ x_2=y_3\ $ and $\ x_3=y_2\ $.
A: 

*

*But 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?

lonza leggiera has nicely shown that $$p\implies c_2.\tag#$$ If we delete the redundant “(or vice versa)” portion of statement $c_1$ and call the revised statement $c_I$ (note that $c_1\iff c_I$), then in fact $$p\iff c_I\quad\text{and}\quad c_I\iff c_2,\tag{*}$$ i.e., the statements $p,$ $c_I$ and $c_2$ are equivalent to one another.
The meat of $(*)$'s proof is the proof of $(\#);$ I shan't belabour the remaining parts.
P.S. Unlike statement $c_1,$ statement $c_2$ does require the “(or vice versa)” clause.
