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Proof step 4 from proof get-firstn (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n < m, n ≤ # x0, m ≤ 1 + # x0, ∀ n, m. n < m ∧ n < # x0 ∧ m ≤ # x0 → firstn(m, x0)[n] = x0[n] ∨ x0 = [] ⊦ firstn(m, a ' + x0)[n] = (a ' + x0)[n]

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ n → (n < m ↔ false); (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' + x ≠ []; (s)% ⊦ a + x = a ' + x; (s)