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Proof step 11 from proof firstoc (previous step)

disj(x, y), disj(y, x), a ∈ x, b ∈ x, ∀ a1, x0. (a1 ∈ y ∨ a1 = b ∨ a1 ∈ y0) ∧ a1 ∈ x0 → (∃ b0, y1, y2. b0 ∈ x0 ∧ y + b ' + y0 = y1 + b0 ' + y2 ∧ disj(x0, y1)), a = b ∨ a ∈ y0 ⊦ a = a0, b ∈ y, a ∈ y, ∃ b0, y1, y2. b0 ∈ x ∧ a0 ' + y + b ' + y0 = y1 + b0 ' + y2 ∧ disj(x, y1), b ∈ x ∧ a0 ' + y + b ' + y0 = (a0 ' + y) + b ' + y0 ∧ disj(x, a0 ' + y)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
disj(y, x), disj(x, y), b ∈ x, a ∈ x, a0 ∈ x, ∀ a1, x0. (a1 ∈ y ∨ a1 = b ∨ a1 ∈ y0) ∧ a1 ∈ x0 → (∃ b0, y1, y2. b0 ∈ x0 ∧ y + b ' + y0 = y1 + b0 ' + y2 ∧ disj(x0, y1)), a = b ∨ a ∈ y0 ⊦ a = a0, a ∈ y, b ∈ y, a0 ∈ y, ∃ b0, y1, y2. b0 ∈ x ∧ a0 ' + y + b ' + y0 = y1 + b0 ' + y2 ∧ disj(x, y1)

Rule argument

Additional information

Used simplifier rules: cons: ⊦ disj(x, y) → (disj(x, a ' + y) ↔ ¬ a ∈ x);%in: ⊦ disj(x, y) ∧ a ∈ x → ¬ a ∈ y;%a: ⊦ (x + y) + z = x + y + z; from specification list