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Proof step 3 from proof firstN-dups-05 (previous step)

n < # x, dups(firstn(n, x)) ⊦ x = [], dups(x)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ n ≤ # x ∧ ¬ dups(x) → ¬ dups(firstn(n, x))

Rule argument

spec: <toplevel> name: firstN-dups-04 seq: ⊦ n ≤ # x ∧ ¬ dups(x) → ¬ dups(firstn(n, x)) subst: [x, n] → [x, n] flags: <all paths>

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < n → ¬ n < m; (s) Used terms: x, n