Define T to be `(\x. x y) z`. Then T and `(\x. x y) z` are syntactically equal, and we're counting them as syntactically equal to `(\z. z y) z` as well, which we will write as:
-> T ≡ `(\x. x y) z` ≡ `(\z. z y) z`
+<pre><code>
+T ≡ (\x. x y) z ≡ (\z. z y) z
+</code></pre>
This: