= Solution
For transitivity of <first-order equality>, take $\phi(v)$ to be $T_1=v$. The <equality elimination rule> applied to $T_2=T_3$ and $\phi(T_2)$ gives
$$
\boxed{\frac{T_2=T_3\qquad T_1=T_2}{T_1=T_3}\ =E.}
$$
For symmetry of <first-order equality>, first obtain $T_1=T_1$ by the <equality introduction rule>. Now take $\psi(v)$ to be $v=T_1$ and substitute using $T_1=T_2$:
$$
\boxed{\frac{T_1=T_2\qquad\dfrac{}{T_1=T_1}\ =I}{T_2=T_1}\ =E.}
$$
These choices of formula show exactly which occurrence is replaced, so neither argument assumes the property it is establishing.
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