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Proof step 17 from proof prefixn-eq-prefixn-app (previous step)

z ≠ [] ∧ # y = n1 ∨ z = [] ∧ # y ≤ n1 ∧ x1 = [] ∨ z = [] ∧ n1 = # y ∧ x1 ≠ [] ⊦ y = (y + z) + x1, n1 ≤ (# y + # z) + # x1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
(# y + # z) + # x1 < n1, z ≠ [] ∧ # y = n1 ∨ z = [] ∧ # y ≤ n1 ∧ x1 = [] ∨ z = [] ∧ n1 = # y ∧ x1 ≠ [] ⊦ z + x1 = []

Rule argument

Additional information

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