Models for predicate logic

Question

Determine whether the following formulae have models:

i) $∃x\,∀y\,(Q(x, x) ∧ ¬Q(x, y))$

ii) $∃x\,∃y\,(P(x) ∧ ¬P(y))$

Not sure if these are right:

i) $D=\{a,b\}$, $Q(a,a)=1$, and $Q(a,b)=0$ (model)

ii) $D=\{a,b\}$, $P(a)=1$, $P(b)=0$ (model)

Answer 1

You (i) is wrong: if $y=x$ you have $Q(x, x) \wedge \neg Q(x,x)$.

Answer 2

You are correct in $ii)$, however $i)$ is not correct as if we choose $x=a$ then the $\forall y$ needs to especially hold for $y=a$ i.e. $Q(a,a)\wedge \neg Q(a,a)$ needs to hold, which it does not in your model.

In fact $i)$ does not have a model which we may show by noticing that it is inconcistent in the following way with a formal proof:

1. Instantiate $\exists x$ with a constant $c$ so we get $\forall y(Q(x,x)\wedge \neg Q(x,c))$ (also called $\exists$ elimination)
2. Instantiate $\forall y$ with the same constant $c$ to get $Q(c,c)\wedge \neg Q(c,c)$ (also called $\forall$ elimination).
3. Split $Q(c,c)\wedge \neg Q(c,c)$ into two sentences $Q(c,c)$ and $\neg Q(c,c)$ and conclude $\bot$, a contradiction.

Thus $\exists x\forall y(Q(x,x)\wedge \neg Q(x,y))\vdash \bot$ and hence by the soundness theorem we conclude that $\exists x\forall y(Q(x,x)\wedge \neg Q(x,y))\models \bot$ and thus the sentence can not have any models, since no model satisfies $\bot$.