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Proof step 3 from proof lastoc (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
a ∈ x0, ∀ a, x0. a ∈ x0 ∧ a ∈ x1 → (∃ b, y, y0. b ∈ x1 ∧ x0 = y + b ' + y0 ∧ disj(x1, y0)), a = a0 ∨ a ∈ x1 ⊦ ∃ b, y, y0. (b = a0 ∨ b ∈ x1) ∧ x0 = y + b ' + y0 ∧ disj(a0 ' + x1, y0)

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)% ⊦ a + x = a ' + x; (s)