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Proof step 18 from proof rec (previous step)

⊦ (∃ a0, x0, y, z. a ' + x = x0 + a0 ' + y + a0 ' + z) ↔ (∃ a0, x0, y, z. x = x0 + a0 ' + y + a0 ' + z), a ∈ x

User applied rule case distinction

And obtained the following new goals

Proof step 19:
∃ a0, x0, y, z. a ' + x = x0 + a0 ' + y + a0 ' + z ⊦ a ∈ x, ∃ a0, x0, y, z. x = x0 + a0 ' + y + a0 ' + z
Proof step 22:
∃ a0, x0, y, z. x = x0 + a0 ' + y + a0 ' + z ⊦ a ∈ x, ∃ a0, x0, y, z. a ' + x = x0 + a0 ' + y + a0 ' + z

Rule argument

pos: right 1

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