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Proof step 8 from proof lastoc (previous step)

∀ a, x0. a ∈ x1 ∧ a ∈ x0 → (∃ 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)

User applied rule exists right

And obtained the following new goals

Proof step 9:
∀ a, x0. a ∈ x1 ∧ a ∈ x0 → (∃ b, y, y0. b ∈ x1 ∧ x0 = y + b ' + y0 ∧ disj(x1, y0)) ⊦ a0 ∈ z, (a0 = a0 ∨ a0 ∈ x1) ∧ y + a0 ' + z = y + a0 ' + z ∧ disj(a0 ' + x1, z), ∃ b, y1, y0. (b = a0 ∨ b ∈ x1) ∧ y + a0 ' + z = y1 + b ' + y0 ∧ disj(a0 ' + x1, y0)

Rule argument

pos: right 2 terms: a0, y, z abort: F discard: F computed: T

Additional information

Used terms: a0, y, z