John

Thanks for getting my head back on LOL. Indeed I have trouble knowing what to assume and where exactly I need to end up. Christopher has a nice answer for **Puzzle 174** so I will write out **Puzzle 173** to practice (hopefully correctly).

> Assume \$$f : X \to Y\$$ is monotone and define \$$\Phi\$$ by

> $\Phi(x,y) \text{ if and only if } f(x) \le y .$

> We want to prove \$$\Phi\$$ is a feasibility relation. So, it suffices to show

> 1. If \$$\Phi(x,y) = \text{true}\$$ and \$$x' \le x\$$ then \$$\Phi(x',y) = \text{true}\$$.
> 2. If \$$\Phi(x,y) = \text{true}\$$ and \$$y \le y'\$$ then \$$\Phi(x,y') = \text{true}\$$.

So we have at our dispense:
$$\Phi(x,y) \text{ if and only if } f(x) \le y$$
$$x' \leq x \;\text{and}\; f(x') \leq f(x)$$
$$y \le y'$$

And we have to show
$$\Phi(x',y') = \text{true}$$

Here we go:
$$\Phi(x,y) = f(x) \le y$$
$$\rightarrow f(x) \le y' \text{ since } y \le y'$$
$$\rightarrow f(x') \le y' \text{ since } x' \leq x \;\text{and}\; f(x') \leq f(x)$$
$$\Phi(x',y') \text{ by definition of } \Phi$$