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Proof step 5 from proof prefixn-eq-empty (previous step)

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

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 1

Additional information

Used simplifier rules: ⊦ firstn(n, x) = [] ↔ n = 0; (s)