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Proof step 2 from proof put-in (previous step)

∀ a, m, b. a ≠ x0[m] ∧ m < # x0 → (a ∈ x0[m, b] ↔ a = b ∨ a ∈ x0) ⊦ a ≠ (a0 + x0)[m] ∧ m < #(a0 + x0) → (a ∈ (a0 + x0)[m, b] ↔ a = b ∨ a ∈ a0 + x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
m ≤ # x0, ∀ a, m, b. a ≠ x0[m] ∧ m < # x0 → (a ∈ x0[m, b] ↔ a = b ∨ a ∈ x0) ⊦ a ∈ (a0 ' + x0)[m, b] ↔ a = b ∨ a = a0 ∨ a ∈ x0, a = (a0 ' + x0)[m]

Rule argument

Additional information

Used simplifier rules: lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; 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%c: ⊦ a + x = a ' + x; from specification list