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Proof step 16 from proof rtc-trans (
previous step
)
p*(b, c), p(a, c0), rtc(n0, c0, b), rtc(n, c0, c) ⊦ ∃ n. rtc(n, a, c), ∃ c0. p(a, c0) ∧ rtc(n, 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