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Proof step 4 from proof firstN-dups-04 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n0 < # x, ∀ x. dups(firstn(n0, x)) ∧ n0 ≤ # x → dups(x), x.head ∈ firstn(n0, x.tail) ∨ dups(firstn(n0, x.tail)) ⊦ x = [], dups(x)

Rule argument

Additional information

Used simplifier rules: firstN-rec: ⊦ firstn(n + 1, x) = x.head + firstn(n, x.tail);% ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ dups(a ' + x) ↔ a ∈ x ∨ dups(x); (s)% ⊦ a + x = a ' + x; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ m +1 = m + 1; (s)