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Proof step 4 from proof rtc-last (previous step)

⊦ (∃ c. p(a, c) ∧ rtc(m, c, b)) ↔ (∃ c. rtc(m, a, c) ∧ p(c, b))

User applied rule case distinction

And obtained the following new goals

Proof step 5:
∃ c. p(a, c) ∧ rtc(m, c, b) ⊦ ∃ c. rtc(m, a, c) ∧ p(c, b)
Proof step 13:
∃ c. rtc(m, a, c) ∧ p(c, b) ⊦ ∃ c. p(a, c) ∧ rtc(m, c, b)

Rule argument

pos: right 1

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