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Proof step 18 from proof prefixn-eq-prefixn-app (previous step)

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

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 2

Additional information

Used simplifier rules: ⊦ [] + x = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ # [] = 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 + n < m; (s)