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Proof step 10 from proof rtc+-unfold-last (previous step)

n0 = 0, p(a, c), rtc(n0, c, b) ⊦ ∃ c. p*(a, c) ∧ p(c, b)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
p(a, b) ⊦ ∃ c. p*(a, c) ∧ p(c, b)

Rule argument

Additional information

Used simplifier rules: rtc-base: ⊦ rtc(0, a, b) ↔ a = b;