I'm studying through Machover's text Set theory, Logic and Their Limitations. How is the follwoing done?
Let $\mathcal L$ be a first-order language with equality but no extra-logical symbols. Let $\sigma$ be a valuation with a universe $\mathit U$ = {$u$, $v$), where $u$ $\neq$ $v$. Let $x^\sigma$ = u for every variable $x$. Let $\Phi$ = {$\phi: \phi^\sigma = T$}. How do I show that $\lnot\forall x(x=y) \in \Phi$, but that there is no term $t$ such that $\lnot(x = y)(x/t) \in \Phi$.
We can use an argument similar to that used for the following post.
We have an interpretation with domain $U = \{ u,v \}$, where the two elements $u$ and $v$ are distinct and we have a valuation $\sigma : \text {Var} \to U$ such that $x^{\sigma}= u$, for every variable $x$.
What are the formulas of the language ? They are built up from atomic formulas using connectives and quantifiers.
But the only atomic formulas are : $x=y$.
They are obviously True in the interpretation [i.e. $(x=y)^{\sigma}= \top$], irrespective that the two variables are distinct or not, because all variables are mapped by $\sigma$ to the same object $u$.
Thus, $[\lnot (x=y)]^{\sigma}= \bot$,
But formula $\forall x (x=y)$ does not hold in the interpretation. Why ?
Consider clause (F4) of 4.6. Basic Semantic Definition [page 154]; we have that $[\forall x (x=y)]^{\sigma}= \bot$ iff $(x=y)^{\sigma(x/a)}= \top$, for some $a \in U$.
By defintion : $x^{\sigma}=y^{\sigma}=u$ but it is enough to choose $v$ as $a$ in the clause above and we have $x^{\sigma(x/v)}=v$.
Thus :
because $u \ne v$.
Conclusion : $[\lnot (\forall x (x=y))]^{\sigma}= \top$.
Now the result is straightforward : the language has no constants and no functions symbols, but only variables. Thus formulas like : $¬(x=y)[x/t]$ are $¬(z=y)$, with $z$ a variable, and we have already shown that they are unsatisfiable by the above interpretation.