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Proof step 7 from proof firstoc (previous step)

a ∈ x, ∀ a, x. a ∈ x1 ∧ a ∈ x → (∃ b, y, y0. b ∈ x ∧ x1 = y + b ' + y0 ∧ disj(x, y)), a = a0 ∨ a ∈ x1 ⊦ a = a0, ∃ b, y, y0. b ∈ x ∧ a0 ' + x1 = y + b ' + y0 ∧ disj(x, y)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
a ∈ x, a ∈ x1, ∀ a, x. a ∈ x1 ∧ a ∈ x → (∃ b, y, y0. b ∈ x ∧ x1 = y + b ' + y0 ∧ disj(x, y)) ⊦ a = a0, ∃ b, y, y0. b ∈ x ∧ a0 ' + x1 = y + b ' + y0 ∧ disj(x, y)

Proof step 7 from proof firstoc (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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