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Proof step 21 from proof rtc-last (previous step)

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

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 2 terms: c1

Additional information

Used terms: c1