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Proof step 11 from proof get-nodups (previous step)

n ≤ # x0, ∀ m0, n0. x0[m0] = x0[n0] ∧ m0 < # x0 ∧ n0 < # x0 → m0 = n0 ∨ x0 = [] ⊦ n = 0, dups(x0), x0[n - 1] ∈ x0

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n, m]

Additional information

Used simplifier rules: get-in: ⊦ m < # x → (x[m] ∈ x ↔ true);% ⊦ m + 1 ≠ 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m