I'm in a mathematical linguistics course, and we are currently learning about the semantics of predicate logic. I was assigned an exercise, and I wanted to make sure that I understood these questions properly.
U(niverse) = {a, b, c}
A = {a, b}
R = {< a,b>, < a,c>, < c,b>, < c,c>}
Given an example model (M1), determine the truth value of the following formulas in M1.
a. R(a,b) ∧ R(b,b)
b. ¬A(c) → R(a,c)
c.∀x [R(x, x)]
d.∀x [R(x, x) ↔ ¬A(x)]
e.∃x∃y∃z [R(x, y) ∧ A(y)∧ R(x, z) ∧ ¬A(z)]
For a)
I got 1 ∧ 0, because R(b, b) is not in R thus the truth value of a) is false. (by semantics of conjunction)
For b)
I got ¬0 → 1, because (c) is not part of A
1 → 1, (by semantics of negation), thus the truth value is true.
For c)
I got all x (a, b, c) there must be a relation where < a,a>, < b,b> and < c,c> occurs
c) is false, because only occurs in R
For d)
I got for all x (a, b, c) there must be a relation where all of < a,a>, < b,b> and < c,c> occurs
For all x (a, b, c), they must be a member of A
0 ↔ ¬0
0 ↔ 1
d) is false, because ∀x [R(x, x) is not equivalent to ¬A(x)]
For e)
Some instance of x, where x is in R < x,y> and x is in relation R
Some instance of y, where y is in relation R < x, y> and is a member of A
Some instance of z, where z is in relation R < x, z> and is a member of A
∃x∃y∃z [R(a, b) ∧ A(b)∧ R(a, c) ∧ ¬A(c)]
∃x∃y∃z [1 ∧ 1 ∧ 1 ∧ ¬0]
∃x∃y∃z [1 ∧ 1 ∧ 1 ∧ 1] (by semantics of negation)
Thus e) is true.
Thank you!
You did a), b). and c) correctly, but you made a mistake in d):
$\forall x [R(x, x) ↔ \neg A(x)]$ says that any object stands in relation $R$ with itself if and only if it does not have property $A$. Now, $a$ and $b$ do have property $A$, but they do not stand in relation $R$ to themselves, so that checks out ok. And $c$ does stand in relation $R$ to itself, but $c$ does indeed not have property $A$. So: it's all good! That is: for every object $x$ we have $R(x,x)$ iff $\neg A(x)$. So, d) is True.
Finally, you have e) correct, but it would be helpful to make explicit what you pick for $x$, $y$, and $z$. So you picked $x=a$, $y=b$, and $z = c$