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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!

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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 $z = c$

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  • $\begingroup$ Thank you so much for taking the time to look over my answers. Your explanations for d) is very helpful in clarifying things! $\endgroup$ – LizJu Jul 15 '17 at 20:22
  • $\begingroup$ @LizJu You're welcome! :) $\endgroup$ – Bram28 Jul 15 '17 at 22:34

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