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Proof step 19 from proof lastoc (previous step)

a ∈ x0, a ∈ x1, a ∈ x0 ∧ a ∈ x1 → (∃ b, y, y0. b ∈ x1 ∧ x0 = y + b ' + y0 ∧ disj(x1, y0)), ∀ a, x0. a ∈ x0 ∧ a ∈ x1 → (∃ b, y, y0. b ∈ x1 ∧ x0 = y + b ' + y0 ∧ disj(x1, y0)) ⊦ ∃ b, y, y0. (b = a0 ∨ b ∈ x1) ∧ x0 = y + b ' + y0 ∧ disj(a0 ' + x1, y0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
disj(y0, x1), disj(x1, y0), b ∈ x1, a ∈ x1, ∀ a, x0. a ∈ x0 ∧ a ∈ x1 → (∃ b, y, y0. b ∈ x1 ∧ x0 = y + b ' + y0 ∧ disj(x1, y0)), a ∈ y ∨ a = b ⊦ a ∈ y0, b ∈ y0, ∃ b0, y1, y2. (b0 = a0 ∨ b0 ∈ x1) ∧ y + b ' + y0 = y1 + b0 ' + y2 ∧ disj(a0 ' + x1, y2)

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% ⊦ ¬ a ∈ x → (a ∈ b ' + x ↔ a = b); (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)