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Proof step 5 from proof firstN-dups-04 (previous step)

x = a + y, n0 < # x, a ∈ firstn(n0, y) ∨ dups(firstn(n0, y)), ∀ x. dups(firstn(n0, x)) ∧ n0 ≤ # x → dups(x) ⊦ x = [], dups(x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: firstN-in: ⊦ ¬ a ∈ x ∧ n ≤ # x → ¬ a ∈ firstn(n, x);%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%rec: ⊦ dups(a ' + x) ↔ ¬ ¬(a ∈ x ∨ dups(x)); from specification list-dup%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%e-01: ⊦ a ' + x ≠ []; from specification list%c: ⊦ a + x = a ' + x; from specification list