Proving symmetry and transitivity of $=$ from axioms

Question

I begin with the following axioms for $=$ (equality):

1. $\forall x (x=x)$
2. $\forall x_1...\forall x_n\forall y_1...\forall y_n ((x_1=y_1 \land ...\land x_n=y_n)\to (p^nx_1...x_n \leftrightarrow p^ny_1...y_n))$ for each n-ary predicate symbol $p^n$ in the first order language
3. $\forall x_1...\forall x_n\forall y_1...\forall y_n ((x_1=y_1 \land ...\land x_n=y_n)\to (f^nx_1...x_n = f^ny_1...y_n))$ for each n-ary function symbol $f^n$ in the first order language

To prove $\forall x\forall y (x=y \to y=x)$ here's what I'm thinking:

• First, I shall use Axiom 2 with $x_1 = x, x_2 = x, y_1 = x, y_2 = y$ to get $\forall x\forall y ((x=x \land x=y) \to (p^2xx \leftrightarrow p^2xy))$. Now, am I free to define $p^2ab$ as $b=a$ for all $a,b$? If yes, then I think I'm done (using Axiom 1, additionally to eliminate occurrences of $x=x$).

To prove $\forall x\forall y\forall z ((x=y\land y=z) \to (x=z))$:

• The obvious choice seems to be axiom 3 this time? $\forall x\forall y\forall z ((x=y\land y=z) \to (f^2xy=f^2yz))$. How do I define $f^2$ to get the result? Are there better ways of proving this?

Answer 1

For transitivity, use the instance of axioms schema 2 where $n=2$ and $p^2t_1t_2\equiv t_1=t_2$, and specialize to $x_1\leftarrow x$, $x_2\leftarrow z$, $y_1\leftarrow y$, $y_2\leftarrow y$ to arrive at $$ (x=y\land z=y)\to (x=z\leftrightarrow y=y)$$ and use axiom 1 and symmetry, and generalize.