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

⊦ prefixn(n, x) = x ↔ n ≥ # 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:
firstn(n, x) = x, n < # x ⊦ x = []

Rule argument

Additional information

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