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Proof step 6 from proof rtc+-unfold-last (previous step)

∀ a, b. rtc(n0, a, b) → n0 = 0 ∨ (∃ c. p*(a, c) ∧ p(c, b)), rtc(n0 +1, a, b) ⊦ n0 +1 = 0, ∃ c. p*(a, c) ∧ p(c, b)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
p(a, c), rtc(n0, c, b), ∀ a, b. rtc(n0, a, b) → n0 = 0 ∨ (∃ c. p*(a, c) ∧ p(c, b)) ⊦ ∃ c. p*(a, c) ∧ p(c, b)

Rule argument

Additional information

Used simplifier rules: rtc-rec: ⊦ rtc(n + 1, a, b) ↔ (∃ c. p(a, c) ∧ rtc(n, c, b));%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2