I've got this predicate symbol:
$(\forall x R(x,y)) \implies (\forall y Q(x,y))$
$R=\{(x,y) \in Q \times Q \hspace{0,2cm}|\hspace{0,2cm} x<y\}$
$Q=\{(x,y) \in Q \times Q \hspace{0,2cm}|\hspace{0,2cm} x=y\}$
Universum is Q
Valuation: $v(x)=-2 \\ v(y)=\frac{3}{5} \\ v(z)=0$
I am not sure, if I understood this rightly.
First part of predicate: $(\forall x R(x,y))$
- $x$ is bound variable
- $y$ is free variable, so I must use valuation $v(y)=\frac{3}{5}$
- in my case it is mean: for all rational numbers applies, that they all are less then $\frac{3}{5}$ -> it is false because for example 10 is from $Q$ and it is greate then $\frac{3}{5}$
Second part of predicate:
$(\forall y Q(x,y))$
- $x$ is free, used $v(x)=-2$
- $y$ is bound
- for all $y$ applies, that are equals to -2 -> false
$(\forall x R(x,y)) \implies (\forall y Q(x,y)) \iff (0 \implies 0) \iff 1$
Predicate symbol has values: TRUE.
Is that correct?