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Proof step 16 from proof rtc-trans (previous step)

Pelem ^*(b, c), Pelem(a, c0), rtc(n0, Pelem)(c0, b), rtc(n, Pelem)(c0, c) ⊦ ∃ n. rtc(n, Pelem)(a, c), ∃ c0. Pelem(a, c0) ∧ rtc(n, Pelem)(c0, c)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 2 terms: c0

Additional information

Used terms: c0