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## Comments

Struggle:In a feasibility relation, why the \( {\text{op}} \) in \( \Phi : X^{\text{op}} \times Y \to \mathbf{Bool} ? \) I really struggled with this. So we have the requirements X, the resources Y and we ask which combinations of elements in X and Y are feasible. Why not \( \Phi : X \times Y \to \mathbf{Bool} \) instead of \( \Phi : X^{\text{op}} \times Y \to \mathbf{Bool} ? \)Resolution:Keep in mind that the feasibility relation \( \Phi \) is supposed to be amonotone function.Allright, let's make an example: Let \( X = \{ Bike, Car, Boat \} \) be a set of vehicles and \( Y = \{ 200$, 1.500$, 1.800$ \} \) a set of money that we have. The feasibility relation \( \Phi : X \times Y \to \mathbf{Bool} \) looks like this:

We want the boat most, but unfortunately we will never be able to buy it. We can buy a car tho if we have 1.500$ or more, and we can buy a bike as soon as we have 1.500$ or more aswell. There are only bikes more expensive than 200$ around. So the feasibility relation Φ is the set { (Bike, 1.500$), (Bike, 1.800$), (Car, 1.500$), (Car, 1.800$) }.

Now we want to think about the feasibility relation as a monotone function between preorders. First in the naive way: \( \Phi : X \times Y \to \mathbf{Bool} \)

Where is the problem? The problem is that there is \( x,x' \in X \) so that \( x \leq x' \) but not \( f(x) \leq f(x')! \) Take for example the pair (Car, 1.500$) and (Boat, 1.500$). \( (Car, 1.500$) \leq (Boat, 1.500$) \) but if we look at where they get mapped at by \( \phi \), we get \( true \leq false\). That is a contraditction to how a monotone function is defined!

When we take the opposite of X, things work out again:

Informally, because of \( X^{\text{op}} \times Y \), everytime we go 'up' or 'right' on the left side of the feasibilty relation, we also go 'up' or 'right' in the right side of the feasibility relation, which wasn't the case with \( X \times Y \).

`**Struggle:** In a feasibility relation, why the \\( {\text{op}} \\) in \\( \Phi : X^{\text{op}} \times Y \to \mathbf{Bool} ? \\) I really struggled with this. So we have the requirements X, the resources Y and we ask which combinations of elements in X and Y are feasible. Why not \\( \Phi : X \times Y \to \mathbf{Bool} \\) instead of \\( \Phi : X^{\text{op}} \times Y \to \mathbf{Bool} ? \\) **Resolution:** Keep in mind that the feasibility relation \\( \Phi \\) is supposed to be a **monotone function**. Allright, let's make an example: Let \\( X = \\{ Bike, Car, Boat \\} \\) be a set of vehicles and \\( Y = \\{ 200$, 1.500$, 1.800$ \\} \\) a set of money that we have. The feasibility relation \\( \Phi : X \times Y \to \mathbf{Bool} \\) looks like this: ![Imgur](https://i.imgur.com/6qn5veq.png) We want the boat most, but unfortunately we will never be able to buy it. We can buy a car tho if we have 1.500$ or more, and we can buy a bike as soon as we have 1.500$ or more aswell. There are only bikes more expensive than 200$ around. So the feasibility relation Φ is the set { (Bike, 1.500$), (Bike, 1.800$), (Car, 1.500$), (Car, 1.800$) }. Now we want to think about the feasibility relation as a monotone function between preorders. First in the naive way: \\( \Phi : X \times Y \to \mathbf{Bool} \\) ![Imgur](https://i.imgur.com/3b5qPOr.png) Where is the problem? The problem is that there is \\( x,x' \in X \\) so that \\( x \leq x' \\) but not \\( f(x) \leq f(x')! \\) Take for example the pair (Car, 1.500$) and (Boat, 1.500$). \\( (Car, 1.500$) \leq (Boat, 1.500$) \\) but if we look at where they get mapped at by \\( \phi \\), we get \\( true \leq false\\). That is a contraditction to how a monotone function is defined! When we take the opposite of X, things work out again: ![Imgur](https://i.imgur.com/cw7UqUq.png) Informally, because of \\( X^{\text{op}} \times Y \\), everytime we go 'up' or 'right' on the left side of the feasibilty relation, we also go 'up' or 'right' in the right side of the feasibility relation, which wasn't the case with \\( X \times Y \\).`