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Proof step 2 from proof lastN-dups (previous step)

∀ n. ¬ dups(x0) ∧ n < # x0 → ¬ dups(lastn(n, x0)) ⊦ ¬ dups(a + x0) ∧ n < #(a + x0) → ¬ dups(lastn(n, a + x0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n ≤ # x0, dups(lastn(n, x0)), ∀ n0. n0 < # x0 → ¬ dups(lastn(n0, x0)) ⊦ a ∈ x0, dups(x0)

Rule argument

Additional information

Used simplifier rules: lastN-append-01: ⊦ n ≤ # y → lastn(n, x + y) = lastn(n, y);%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%a: ⊦ (x + y) + z = x + y + z; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%c: ⊦ a + x = a ' + x; from specification list