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

∀ a, m, b. 1 < #oc(a, x0) ∧ m < # x0 → a ∈ x0[m, b] ⊦ 1 < #oc(a, a0 + x0) ∧ m < #(a0 + x0) → a ∈ (a0 + x0)[m, b]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
m ≤ # x0, ∀ a1, m0, b0. a1 ∈ x0 ∧ #oc(a1, x0) ≠ 1 ∧ m0 < # x0 → a1 ∈ x0[m0, b0] ⊦ #oc(a, a0 ') + #oc(a, x0) = 0, a ∈ (a0 ' + x0)[m, b], a ≠ a0 ∧ #oc(a, x0) = 1, ¬ a ∈ x0 ∧ #oc(a, a0 ') = 1

Rule argument

Additional information

Used simplifier rules: ocappend: ⊦ #oc(a, x + y) = #oc(a, x) + #oc(a, y);%oc-zero: ⊦ #oc(a, x) = 0 ↔ ¬ a ∈ x;%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%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%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%in-17: ⊦ a ∈ b ' ↔ a = b; from specification list%one: ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1; from specification nat-basic2%one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list