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Proof step 7 from proof in-rec (previous step)

⊦ (∃ y, z. b ' + x = y + a ' + z) ↔ a = b ∨ (∃ y, z. x = y + a ' + z)

User applied rule case distinction

And obtained the following new goals

Proof step 8:
∃ y, z. b ' + x = y + a ' + z ⊦ a = b ∨ (∃ y, z. x = y + a ' + z)
Proof step 12:
a = b ∨ (∃ y, z. x = y + a ' + z) ⊦ ∃ y, z. b ' + x = y + a ' + z

Rule argument

pos: right 1

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