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Proof step 2 from proof in-prefixn-at-smaller (previous step)

m < n, m < # x ⊦ prefixn(n, x)[m] = x[m]

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:
m < n, m < # x ⊦ (n ≤ # x ⊃ firstn(n, x);x)[m] = x[m], x = []

Rule argument

Additional information

Used simplifier rules: prefixn: ⊦ prefixn(n, x) = (n ≤ # x ⊃ firstn(n, x);x);% ⊦ # x = 0 ↔ x = []; (s)% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic