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Proof step 7 from proof in-put (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
m ≤ # x0, a ∈ x0, ∀ a1, m0, b0. a1 ∈ x0 ∧ #oc(a1, x0) ≠ 1 ∧ m0 < # x0 → a1 ∈ x0[m0, b0] ⊦ a = a0, #oc(a, x0) = 1, a ∈ (a0 ' + x0)[m, b]

Rule argument

Additional information

Used simplifier rules: one-01: ⊦ a ≠ b → #oc(a, b ') = 0;%oc-zero: ⊦ #oc(a, x) = 0 ↔ ¬ a ∈ x;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-base: ⊦ n + 0 = n; from specification nat-basic2