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Proof step 8 from proof put-in (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
m0 < # x0, ∀ m, a, b. m < # x0 → (a ∈ x0[m, b] ↔ a = b ∨ a ∈ x0) ∨ a = x0[m] ⊦ a ∈ x0[m0, b] ↔ a = b ∨ a ∈ x0, a = x0[m0], a = a0, x0 = []

Rule argument

Additional information

Used simplifier rules: get-succ: ⊦ (a ' + x)[n + 1] = x[n];% ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)