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Proof step 17 from proof dups-def (previous step)

a = (a ' + x0)[n], n ≤ # x0 ⊦ 0 = n, dups(x0), a ∈ x0, ∃ m, n. m < # x0 ∧ n < # x0 ∧ m ≠ n ∧ x0[m] = x0[n], 0 - 1 < # x0 ∧ n - 1 < # x0 ∧ 0 - 1 ≠ n - 1 ∧ x0[0 - 1] = x0[n - 1]

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, m0]

Additional information

Used simplifier rules: get-succ: ⊦ (a ' + x)[n + 1] = x[n];%get-in: ⊦ m < # x → (x[m] ∈ x ↔ true);%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ m + 1 ≠ 0; (s)% ⊦ m + 1 ≤ n ↔ m < n; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ # x = 0 ↔ x = []; (s)%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m0