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

a = (a ' + x0)[n], n ≤ # x0 ⊦ 0 = n, a ∈ x0, dups(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 inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n, m0] flags:

Additional information

Used simplifier rules: get-succ: ⊦ (a ' + x)[n + 1] = x[n];%get-in: ⊦ m < # x → (¬ x[m] ∈ x ↔ false);%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%l: ⊦ # x = 0 ↔ x = []; from specification list-data%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m0