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Proof step 13 from proof rtc-last (previous step)

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

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: c

Additional information

Used terms: c