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Proof step 7 from proof firstoc (previous step)

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

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 4 terms: b, a0 ' + y, y0

Additional information

Used simplifier rules: cons: ⊦ disj(x, y) → (disj(x, a ' + y) ↔ ¬ a ∈ x);%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: b, a0 ' + y, y0