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Proof step 10 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) ⊦ a ∈ (a0 ' + x0)[m, a0][0, x0[m - 1]] ↔ a ∈ a0 ' + 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: 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 ↔ 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% ⊦ a + x = a ' + x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ m + 1 ≤ n ↔ m < n; (s)%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m, m1