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Proof step 9 from proof prefixn-eq-prefixn-app (previous step)

x0 = (n1 ≤ # x0 + # x1 ⊃ firstn(n1, x0 + x1);x0 + x1), # x0 < n0 ⊦ # x0 ≤ n1 ∧ x1 = [], n1 = # x0 ∧ x1 ≠ []

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 10:
x0 = firstn(n1, x0 + x1), # x0 < n0, n1 ≤ # x0 + # x1 ⊦ n1 = # x0 ∧ x1 ≠ [], # x0 ≤ n1 ∧ x1 = []

Rule argument

pos: left 1

Additional information

Used simplifier rules: ⊦ x = x + y ↔ y = []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ m ≤ n ↔ ¬ n < m; (ss)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic% ⊦ # x = 0 ↔ x = []; (s)% ⊦ m < n → n ≠ 0; (s)% ⊦ m +1 = m + 1; (s)% ⊦ # [] = 0; (s)% ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + m0 < n → ¬ n + n0 < m; (s)