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

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

System applied rule normalize

And obtained the following new goals

Proof step 2:
n1 ≤ # x, n0 ≤ # x ⊦ prefixn(n0, x) = prefixn(n1, x) ↔ n0 = n1

Rule argument

Additional information