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Proof step 6 from proof firstoc (previous step)

a ∈ x, a ∈ x1, a ∈ x ∧ a ∈ x1 → (∃ b, y, y0. b ∈ x ∧ x1 = y + b ' + y0 ∧ disj(x, y)), ∀ a, x. a ∈ x ∧ a ∈ x1 → (∃ b, y, y0. b ∈ x ∧ x1 = y + b ' + y0 ∧ disj(x, y)) ⊦ a0 ∈ x, ∃ 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 7:
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)

Rule argument

Additional information

Used simplifier rules: in: ⊦ disj(x, y) ∧ a ∈ x → ¬ a ∈ y;%sym: ⊦ disj(x, y) → disj(y, x);%a: ⊦ (x + y) + z = x + y + z; from specification list%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list