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Proof step 4 from proof prefixn-shorter (previous step)

prefixn(n + 1, []) = prefixn(n + 1, x1) ⊦ prefixn(n, []) = prefixn(n, x1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
[] = prefixn(n + 1, x1) ⊦ n = 0, x1 = []

Rule argument

Additional information

Used simplifier rules: ⊦ prefixn(n, x) = [] ↔ n = 0 ∨ x = []; (ss)% ⊦ prefixn(n, []) = []; (s)