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Proof step 9 from proof put-in-01 (previous step)

(a0 ' + x0)[m] = a0, m ≤ # x0, ∀ m, a, b. m < # x0 → (a ∈ x0[m, b] ↔ a = b ∨ a ∈ x0 -1l x0[m]) ⊦ a = a0 ∨ a ∈ x0[m - 1, b] ↔ a = b ∨ a ∈ x0, m = 0

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] → [m, m1]

Additional information

Used simplifier rules: get-succ: ⊦ (a ' + x)[n + 1] = x[n];%put-in: ⊦ a ≠ x[m] ∧ m < # x → (a ∈ x[m, b] ↔ a = b ∨ a ∈ x);%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: m, m1