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Proof step 3 from proof prefixn-eq-empty (previous step)

⊦ (n ≤ # x ⊃ firstn(n, x);x) = [] ↔ n = 0 ∨ x = []

User applied rule case distinction

And obtained the following new goals

Proof step 4:
(n ≤ # x ⊃ firstn(n, x);x) = [] ⊦ n = 0, x = []

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ 0 ≤ m; (s)% ⊦ firstn(0, x) = []; (s)% ⊦ # [] = 0; (s)% ⊦ m ≤ 0 ↔ m = 0; (s)