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Proof step 3 from proof firstoc (previous step)

a ∈ x, ∀ a, x. a ∈ x ∧ a ∈ x1 → (∃ 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)

User applied rule exists right

And obtained the following new goals

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

Rule argument

pos: right 1 terms: a0, [], x1 abort: F discard: F computed: F

Additional information

Used terms: a0, [], x1