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Proof step 5 from proof split-first (previous step)

n ≤ 1 + # x0, n - 1 ≤ # x0 → x0 = y + z ∧ # y = n - 1 ⊦ ∃ y, z. a ' + x0 = y + z ∧ # y = n

User applied rule case distinction

And obtained the following new goals

Proof step 6:
# y = n - 1, x0 = y + z, n ≤ 1 + # x0 ⊦ ∃ y, z. a ' + x0 = y + z ∧ # y = n

Rule argument

pos: left 2

Additional information

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