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Proof step 19 from proof rtc-last (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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