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

⊦ # x ≤ # prefixn(n, x) ↔ 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)

Rule argument

Additional information

Used simplifier rules: prefixn: ⊦ prefixn(n, x) = (n ≤ # x ⊃ firstn(n, x);x);% ⊦ m < n → (n ≤ m ↔ false); (s)% ⊦ n < # x → # firstn(n, x) = n; (s)% ⊦ m < n → m ≤ n; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m ≤ m; (s)% ⊦ firstn(# x, x) = x; (s)% ⊦ m ≤ n → (n ≤ m ↔ m = n); (s)% ⊦ n ≥ m ↔ m ≤ n; (s)