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Proof step 3 from proof firstN-dups-04 (previous step)

dups(firstn(n, x)), n ≤ # x ⊦ dups(x)

User applied rule structural induction

And obtained the following new goals

Proof step 4:
dups(firstn(n0 +1, x)), n0 +1 ≤ # x, ∀ x. dups(firstn(n0, x)) ∧ n0 ≤ # x → dups(x) ⊦ dups(x)

Rule argument

indvar: n

Additional information

Used simplifier rules: firstN-zero: ⊦ firstn(0, x) = [];%nat freely generated by 0, +1;;% ⊦ 0 ≤ m; (s)% ⊦ ¬ dups([]); (s)