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Proof step 3 from proof rtc-last (previous step)

n = m + 1 ⊦ rtc(n, a, b) ↔ (∃ c. rtc(m, a, c) ∧ p(c, b)), n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ (∃ c. p(a, c) ∧ rtc(m, c, b)) ↔ (∃ c. rtc(m, 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));%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2