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Proof step 18 from proof put-sw-in (previous step)

m = m1 + 1, 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, x0[m1]] ↔ a ∈ a0 ' + x0, m = 0, x0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
m0 < # x0, m1 < # x0, ∀ m, n, a. m < # x0 ∧ n < # x0 → (a ∈ x0[m, x0[n]][n, x0[m]] ↔ a ∈ x0) ⊦ a ∈ x0[m1, x0[m0]][m0, x0[m1]] ↔ a ∈ x0, a = a0, x0 = []

Rule argument

Additional information

Used simplifier rules: put-succ: ⊦ (b ' + x)[n + 1, a] = b + x[n, a];% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)