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Proof step 9 from proof rtc-last (previous step)

p(a, c), p(c, c0), rtc(m0, c0, b), p(c, c0) ∧ rtc(m0, c0, b) → (∃ c0. rtc(m0, c, c0) ∧ p(c0, b)) ⊦ ∃ c1. (∃ c. p(a, c) ∧ rtc(m0, c, c1)) ∧ p(c1, b)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information