firstOrderLogic - ∃xy.(P(x) ∧ ¬P(y))

Question

I'm trying to make a model to satisfy the following formula

∀xyz.((R(x,y) ∧ R(y,z)) → R(z,z))
∃xy.(P(x) ∧ ¬P(y))
∀x ∃y.(R(x,y) ∧ x ≠ y)

I made this.

D= {Susan, George, Bob, Emma, Sara, Ryan, Jacob}
I(P) = {Susan, George, Bob, Emma}
I(R) = {〈George, Emma〉, 〈Emma, Bob〉, 〈Bob, Bob〉, 〈Bob, Emma〉, 〈Emma, Emma〉, 〈Susan, Jacob〉}

But I don't know how to make P be satisfied!! and what does "∃xy.(P(x) ∧ ¬P(y))" mean exactly!

Thanks in advance

Answer 1

$\exists xy.(P(x) \wedge \lnot P(y))$ means there are two items in your universe such that $P$ is true about one and false about the other. Your set $I(P)$ looks like the items for which $P$ is true, so you have $P(Susan)$ and $\lnot P(Ryan)$ and the statement is satisfied in your model.

Your model does not satisfy the first and third statements. For the third, there is no pair $R(Jacob, y)$ with $y$ different from Jacob.

Answer 2

I'm not sure what you're trying to do with the names, but on first glance the third formula isn't satisfied: taking $x$ to be Sara, Ryan, or Jacob makes it false.

In general wouldn't it be easier to use $\mathbb{Z}$ as your universe? Interpret the binary relation as the standard partial order and the unary relation (that's what $P(x)$ stands for - essentially just a subset of the chosen universe) as the property of being odd. Then, written in more convenient notation, our formulas become:

$$\forall x,y,z\text{ }\big((x\leq y\text{ } \land y\leq z)\to z\leq z\big) \\ \exists x,y\text{ } (Odd(x) \text{ }\land \neg Odd(y)) \\ \forall x \exists y\text{ } (x\leq y\text{ }\land x\neq y)$$ I think these are easily seen to be satisfied. Hopefully the odd-not odd example clarifies what confusion you had with the second statement.