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

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

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] → [m, m1] flags:

Additional information

Used simplifier rules: put-succ: ⊦ (b ' + x)[n + 1, a] = b + x[n, a];%put-zero: ⊦ (b ' + x)[0, a] = a + x;%put-in: ⊦ a ≠ x[m] ∧ m < # x → (a ∈ x[m, b] ↔ a = b ∨ a ∈ x);%get-in: ⊦ m < # x → (¬ x[m] ∈ x ↔ false);%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%c: ⊦ a + x = a ' + x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m, m1