Satisfiability of predicate logic formulas

Question

I have to find worlds that satisfy / do not satisfy the formulas

My W is the notation for my worlds U:= Universe

Now i am at the fourth and can literally not find a satisfying world but my tutors told me as a hint that there exists one

1. $\psi_{1}=\forall x\left(\exists x^{\prime}\left(R\left(x, x^{\prime}\right) \wedge \neg R\left(x^{\prime}, x\right)\right)\right)$

$W_1$

$S= K^2$

$U = \mathbb{N}$

R=(x,x')| x > x'

$\alpha(x) = 2$ , $\alpha(x')=1$

$W_1 \models \varphi_1$

$W_2$

$S= K^2$\

$U = \mathbb{N}$

R=(x,x')| x = x'

$\alpha(x) = 2$ , $\alpha(x')=2$

$W_2 \not\models \varphi_1$

2. $\psi_{2}=\exists x(R(x, x)) \wedge \forall x\left(\forall x^{\prime}\left(\neg R\left(x, x^{\prime}\right) \leftrightarrow R\left(x^{\prime}, x\right)\right)\right)$

$W_1$ $S= K^2$ $U = \mathbb{N}$ R=(x,x')| x = x' $\alpha(x) = 2$ , $\alpha(x')=2$ $W_1 \not\models \varphi_1$

It's a contradiction.\

3. $\psi_{3}=\exists x\left(\forall x^{\prime \prime}\left(R\left(x, x^{\prime \prime}\right) \vee x=x^{\prime \prime} \vee \exists x^{\prime}\left(x=x^{\prime}\right)\right)\right)$

$W_1$

$S= K^2$

$U = \{ 1 \}$

R=(x,x'')| x = x''

$\alpha(x) = 1$ , $\alpha(x')=1$

$W_1 \models \varphi_1$

Tautology

4.$\psi_{4}=\exists x\left(\exists x^{\prime}\left(\exists x^{\prime \prime}\left(\left(x=x^{\prime}\right) \wedge \neg\left(x=x^{\prime \prime}\right) \wedge\left(R\left(x, x^{\prime}\right) \rightarrow\left(x^{\prime}=x^{\prime \prime}\right)\right)\right)\right)\right)$

Answer 1

Just have a world with two different objects $a$ and $b$ and for which $R(a,a)$ does not hold.

Then the statement is true, since for $x$ and $x'$ you can pick $a$, and for $x''$ you pick $b$.... specifically note that since $R(a,a)$ does not hold, you have that $R(x,x')$ is False, and hence that $R(x,x') \to x'=x''$ is True.