Michael wrote:

> **Puzzle 173.** If we let \\(y = f(x')\\) then \\(f(x) \leq y\\) sets up the following inequalities by definition of a monotone function :

> $$x \leq x' \;\text{and}\; f(x) \leq f(x')$$

> By definition of a feasibility relation the above inequality implies :

> $$\Phi(x,f(x)) \;\text{implies}\; \Phi(x',f(x'))$$

> So...

> $$\text{If} \; \Phi(x,f(x)) = \text{true and} \; f(x) \leq y, \text{then} \; \Phi(x,y) = \text{true}$$

I have trouble following what you're doing here. The question asks you to _assume_ \\(\Phi\\) is defined by

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

where \\(f\\) is monotone, and _prove_ that \\(\Phi\\) is a feasibility relation.

So, it makes me very nervous when you say "by the definition of a feasibility relation the above inequality implies...." That sounds like you are _assuming_ \\(\Phi\\) is a feasibility relation and deducing things from this. But this is what you're trying to _prove_.

It's quite possible you have a good idea and I'm getting stumped by the way you're explaining it. Here's how a mathematician would start an answer to this puzzle:

> 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}\\).

In other words: state what you're assuming and what you're trying to show. Then use the assumptions to get what you're trying to show.

Another fine approach starts with the original definition of feasibility relation:

> 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 that if \\(\Phi(x,y) = \text{true}\\) and \\(x' \le x\\) and \\(y \le y'\\), then \\(\Phi(x',y') = \text{true}\\).