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Proof step 1 from proof rtc+-unfold-last (conclusion)

⊦ p+(a, b) ↔ (∃ c. p*(a, c) ∧ p(c, b))

User applied rule case distinction

And obtained the following new goals

Proof step 2:
p+(a, b) ⊦ ∃ c. p*(a, c) ∧ p(c, b)
Proof step 17:
∃ c. p*(a, c) ∧ p(c, b) ⊦ p+(a, b)

Rule argument

pos: right 1

Additional information