# Satisfiability of predicate logic formulas

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'

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

$$W_2 \not\models \varphi_1$$

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

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

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

$$W_1$$

$$S= K^2$$

$$U = \{ 1 \}$$

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

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

$$W_1 \models \varphi_1$$

Tautology

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

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.