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Proof step 11 from proof rtc-last (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information