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Proof step 23 from proof prefixn-eq-prefixn-app (previous step)

x0 + x1 = y + z, # x0 < n0, # y ≤ # x0 + # x1, # x0 ≤ # y ∧ x1 = [] ∨ # y = # x0 ∧ x1 ≠ [] ⊦ x0 = y

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 4

Additional information

Used simplifier rules: ⊦ # x = # z → (x + y = z + y0 ↔ x = z ∧ y = y0); (ws)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ x + [] = x; (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 ≤ n → (n ≤ m ↔ m = n); (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic% ⊦ #(x + y) = # x + # y; (s)% ⊦ m = m + n ↔ n = 0; (s)% ⊦ # x = 0 ↔ x = []; (s)% ⊦ x = x + y ↔ y = []; (s)