Exercise 7, § 5

Show that if the scheme $(R \Leftrightarrow S) \Rightarrow (\tau_x(R) = \tau_x(S))$ provides implicit axioms in $\mathscr{T}$, then $x = y$ is a theorem in $\mathscr{T}$. (Take $R$ to be the relation $x = x$, $S$ the relation $x = y$, and then substitute $x$ for $y$ in the axiom thus obtained.) [^1]

[^1]: We shall see that in the theory of sets $(\exists x)(\exists v)(x \neq y)$ is a theorem (Chapter II, § 1, Exercise 2).