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Proof step 3 from proof get-firstn (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n ≤ # x0, m ≤ # x0 + 1, n < m, ∀ n, m. n < m ∧ m ≤ # x0 ∧ n < # x0 → firstn(m, x0)[n] = x0[n] ∨ m = 0 ∨ x0 = [] ⊦ firstn(m, a ' + x0)[n] = (a ' + x0)[n], m = 0

Rule argument

Additional information

Used simplifier rules: le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%e-01: ⊦ a ' + x ≠ []; from specification list%c: ⊦ a + x = a ' + x; from specification list