Lemma UAL-5. Identity for two 5-ary Steiner laws. The type is ⟨5, 5, 0⟩.

\{ \begin{matrix}f(u, v, w, x, y) = f(u, v, w, y, x) \\ f(e, e, e, e, e) = e \\ f(u, v, w, x, f(u, v, w, x, y)) = y \\ g(u, v, w, x, y) = g(u, v, w, y, x) \\ g(e, e, e, e, e) = e \\ g(u, v, w, x, g(u, v, w, x, y)) = y \end{matrix} \} =(gL)⇒

{f(x_{1}, x_{2}, x_{3}, y, g(x_{4}, x_{5}, x_{6}, y, x_{7})) = f(x_{1}, x_{2}, x_{3}, z, g(x_{4}, x_{5}, x_{6}, z, x_{7}))}.

Proof (found by Otter 3.0.4 on gyro at 22.00 seconds).

2 f(u, v, w, x, y) = f(u, v, w, y, x) 5 g(u, v, w, x, y) = g(u, v, w, y, x) 6 g(e, e, e, e, e) = e 7 g(u, v, w, x, g(u, v, w, x, y)) = y 8 f(A_{1}, A_{2}, A_{3}, B, g(A_{4}, A_{5}, A_{6}, B, A_{7})) ≠

f(A_{1}, A_{2}, A_{3}, C, g(A_{4}, A_{5}, A_{6}, C, A_{7}))

10 f(x, y, z, u, g(e, e, e, e, e)) = f(x, y, z, e, u)  [6 → 2]

14 f(x, y, z, e, g(u, v, w, v_{6}, g(u, v, w, v_{6}, v_{7}))) = f(x, y, z, v_{7}, g(e, e, e, e, e))

[7 → 10, flip]

20 f(x, y, z, e, g(u, v, w, v_{6}, g(u, v, w, v_{7}, v_{6}))) = f(x, y, z, v_{7}, g(e, e, e, e, e))

[5 → 14]

24 f(x, y, z, e, g(u, v, w, v_{6}, g(u, v, w, e, v_{7}))) = f(x, y, z, v_{7}, g(e, e, e, v_{6}, e))

[(gL) 14]

45 f(x, y, z, e, g(u, v, w, e, v_{6})) = f(x, y, z, v_{7}, g(u, v, w, v_{7}, v_{6}))

[20 → 24 :(gL) :(gL) :(gL) :(gL)]

52 f(x, y, z, u, g(v, w, v_{6}, u, v_{7})) = f(x, y, z, v_{8}, g(v, w, v_{6}, v_{8}, v_{7})) [45 → 45]
53 □  [52,8]