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Proof step 5 from proof in-put (previous step)

a = a0, 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 ∈ x0 ∧ #oc(a, a0 ') = 1, a ≠ a0 ∧ #oc(a, x0) = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
a0 ∈ x0, m ≤ # x0, ∀ a1, m0, b0. a1 ∈ x0 ∧ #oc(a1, x0) ≠ 1 ∧ m0 < # x0 → a1 ∈ x0[m0, b0] ⊦ a0 ∈ (a0 ' + x0)[m, b]

Rule argument

Additional information

Used simplifier rules: one: ⊦ #oc(a, a ') = 1;%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)