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Proof step 2 from proof get-from-01 (previous step)

m + m0 < # x ⊦ (x from m)[m0] = x[m + m0], x = []

User applied rule structural induction

And obtained the following new goals

Proof step 3:
m + m0 < #(a + x0), ∀ m, m0. m + m0 < # x0 → (x0 from m)[m0] = x0[m + m0] ∨ x0 = [] ⊦ a + x0 = [], (a + x0 from m)[m0] = (a + x0)[m + m0]

Rule argument

indvar: x

Additional information

Used simplifier rules: list freely generated by [], +;;