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Proof step 3 from proof in-put (previous step)

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

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 4:
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 ≠ a0 ∧ #oc(a, x0) = 1, ¬ a ∈ x0 ∧ #oc(a, a0 ') = 1
Proof step 7:
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

Rule argument

fma: a = a0

Additional information

Used formulas: a = a0