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Proof step 12 from proof put-sw-in (previous step)

n = m0 + 1, m ≤ # x0, n ≤ # x0, ∀ m, n, a. m < # x0 ∧ n < # x0 → (a ∈ x0[m, x0[n]][n, x0[m]] ↔ a ∈ x0) ⊦ n = 0, a ∈ (a0 ' + x0)[m, x0[m0]][n, (a0 ' + x0)[m]] ↔ a = a0 ∨ a ∈ x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
m0 < # x0, m ≤ # x0, ∀ m, n, a. m < # x0 ∧ n < # x0 → (a ∈ x0[m, x0[n]][n, x0[m]] ↔ a ∈ x0) ⊦ a ∈ (a0 ' + x0)[m, x0[m0]][m0 + 1, (a0 ' + x0)[m]] ↔ a = a0 ∨ a ∈ x0, x0 = []

Rule argument

Additional information

Used simplifier rules: l: ⊦ # x = 0 ↔ x = []; from specification list-data%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2