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Proof step 6 from proof get-firstn (previous step)

n ≤ # x0, m ≤ # x0 + 1, n < m, ∀ n0, m0. n0 < m0 ∧ m0 ≤ # x0 ∧ n0 < # x0 → firstn(m0, x0)[n0] = x0[n0] ∨ m0 = 0 ⊦ x0 = [], firstn(m, a ' + x0)[n] = (a ' + x0)[n], m = 0

User applied rule structural induction

And obtained the following new goals

Proof step 7:
∀ n, x0, a. n ≤ # x0 ∧ m0 ≤ # x0 + 1 ∧ n < m0 ∧ (∀ n0, m0. n0 < m0 ∧ m0 ≤ # x0 ∧ n0 < # x0 → firstn(m0, x0)[n0] = x0[n0] ∨ m0 = 0) → x0 = [] ∨ firstn(m0, a ' + x0)[n] = (a ' + x0)[n] ∨ m0 = 0, n ≤ # x0, m0 +1 ≤ # x0 + 1, n < m0 +1, ∀ n0, m0. n0 < m0 ∧ m0 ≤ # x0 ∧ n0 < # x0 → firstn(m0, x0)[n0] = x0[n0] ∨ m0 = 0 ⊦ x0 = [], firstn(m0 +1, a ' + x0)[n] = (a ' + x0)[n], m0 +1 = 0

Rule argument

indvar: m

Additional information

Used simplifier rules: nat freely generated by 0, +1;;