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Proof step 15 from proof lastoc (previous step)

a ∈ z, a ∈ x1, a ∈ z ∧ a ∈ x1 → (∃ b, y, y0. b ∈ x1 ∧ z = y + b ' + y0 ∧ disj(x1, y0)), ∀ a, x0. a ∈ x0 ∧ a ∈ x1 → (∃ b, y, y0. b ∈ x1 ∧ x0 = y + b ' + y0 ∧ disj(x1, y0)) ⊦ a0 ∈ z, ∃ b, y1, y0. (b = a0 ∨ b ∈ x1) ∧ y + a0 ' + z = y1 + b ' + y0 ∧ disj(a0 ' + x1, y0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
disj(y1, x1), disj(x1, y1), b ∈ x1, a ∈ x1, ∀ a, x0. a ∈ x0 ∧ a ∈ x1 → (∃ b, y, y0. b ∈ x1 ∧ x0 = y + b ' + y0 ∧ disj(x1, y0)), a ∈ y0 ∨ a = b ⊦ a0 = b, a ∈ y1, b ∈ y1, a0 ∈ y0, a0 ∈ y1, ∃ b0, y3, y2. (b0 = a0 ∨ b0 ∈ x1) ∧ y + a0 ' + y0 + b ' + y1 = y3 + 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 ∈ b ' ↔ a = b; (s)%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)