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Proof step 7 from proof get-firstn (previous step)

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

User applied rule structural induction

And obtained the following new goals

Proof step 8:
n < m0 +1, m0 +1 ≤ 1 + # x0, n ≤ # x0, ∀ n0, m0. n0 < m0 ∧ n0 < # x0 ∧ m0 ≤ # x0 → firstn(m0, x0)[n0] = x0[n0], ∀ n, x0, a. n < m0 ∧ n ≤ # x0 ∧ m0 ≤ 1 + # x0 ∧ (∀ n0, m0. n0 < m0 ∧ n0 < # x0 ∧ m0 ≤ # x0 → firstn(m0, x0)[n0] = x0[n0]) → firstn(m0, a ' + x0)[n] = (a ' + x0)[n] ∨ x0 = [] ⊦ firstn(m0 +1, a ' + x0)[n] = (a ' + x0)[n], x0 = []

Rule argument

indvar: m

Additional information

Used simplifier rules: firstN-zero: ⊦ firstn(0, x) = [];%nat freely generated by 0, +1;;% ⊦ 0 ≤ m; (s)% ⊦ ¬ n < 0; (s)