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

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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 5:
n ≤ # x0, m ≤ # x0 + 1, n < m, ∀ n, m. n < m ∧ m ≤ # x0 ∧ n < # x0 → firstn(m, x0)[n] = x0[n] ∨ m = 0 ∨ x0 = [] ⊦ x0 = [], firstn(m, a ' + x0)[n] = (a ' + x0)[n], m = 0

Rule argument

fma: x0 = []

Additional information

Used simplifier rules: firstn-one-02: ⊦ firstn(1, a ') = a ';%one-03: ⊦ a '[0] = a;%app: ⊦ x + [] = x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%size-01: ⊦ # [] = 0; from specification list-data%lez-01: ⊦ m ≤ 0 ↔ m = 0; from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%one-02: ⊦ m ≠ 0 → (m ≤ 1 ↔ m = 1); from specification nat%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1 Used formulas: x0 = []