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Proof step 2 from proof prefixn-le-eq (previous step)

n1 ≤ # x, n0 ≤ # x ⊦ prefixn(n0, x) = prefixn(n1, x) ↔ n0 = n1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ prefixn(n, x) = (n ≤ # x ⊃ firstn(n, x);x)
Proof step 4:
firstn(n0, x) = firstn(n1, x), n0 ≤ # x, n1 ≤ # x ⊦ n0 = n1

Rule argument

Additional information

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