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Proof step 10 from proof in-put (previous step)

m ≠ 0, 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 -1, b]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
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, m = 0, a ∈ x0[m - 1, b]

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list%in-01: ⊦ a ≠ b → (a ∈ b ' + x ↔ a ∈ x); from specification list%pred-03: ⊦ n -1 = n - 1; from specification nat