Is every non-increasing function not increasing? I want to prove or disprove my hypothesis:
Let $f:(a,b)\rightarrow R$ be non-increasing function, then it is not increasing.

Definitions: 
$$\text{Function is increasing iff $x&lt;y\Rightarrow f(x)&lt;f(y)$}\tag{1}$$

$$\text{Function is non-increasing iff $x&lt;y\Rightarrow f(x)\geq f(y)$}\tag{2}$$
So, my idea was to just write the theorem as it is:
$$\{\forall x,y\in(a,b):x&lt;y\Rightarrow f(x)\geq f(y)\}\Rightarrow \{\forall x,y\in(a,b):x&lt;y\Rightarrow f(x)&lt; f(y)\}$$
And make use of truth values of $x&lt;y,f(x)&lt;f(y)$ but I don't get a tautology, I couldn't make use of the quantifiers.

Secondly I would be interested in disproving the converse statement that any not inreasing function is non-increasing. As a counter example i would give $x\mapsto \sin{x}$ on $(0,2\pi)$
Non-increasing:

$$\forall\,x, y: x&lt;y:f(x)\ge f(y).$$

Not increasing:

$$\overline{\forall\,x,y:x&lt;y:f(x)&lt;f(y)}\iff \exists\,x,y:x&lt;y:f(x)\ge f(y).$$

Obviously,

$$\forall\,x, y: x&lt;y:f(x)\ge f(y)\implies \exists\,x,y:x&lt;y:f(x)\ge f(y).$$

Notice the change of the quantifier.
Hint:  By trichotomy $a &lt; b \iff a \not \ge b$.