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Proof step 1 from proof prefixn-len-le-n (conclusion)

⊦ # prefixn(n, x) ≤ n

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:
n < # (n ≤ # x ⊃ firstn(n, x);x) ⊦ n = 0

Rule argument

Additional information

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