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Proof step 3 from proof firstoc (previous step)

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

Rule argument

Additional information

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