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Proof step 1 from proof prefixn-eq-empty (conclusion)

⊦ prefixn(n, x) = [] ↔ n = 0 ∨ x = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ prefixn(n, x) = (n ≤ # x ⊃ firstn(n, x);x)
Proof step 3:
⊦ (n ≤ # x ⊃ firstn(n, x);x) = [] ↔ n = 0 ∨ x = []

Rule argument

Additional information

Used simplifier rules: prefixn: ⊦ prefixn(n, x) = (n ≤ # x ⊃ firstn(n, x);x);