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Proof step 2 from proof rtc-last (previous step)

⊦ rtc(n, Pelem)(a, b) ↔ (∃ c. rtc(n - 1, Pelem)(a, c) ∧ Pelem(c, b)), n = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
n = m + 1 ⊦ rtc(n, Pelem)(a, b) ↔ (∃ c. rtc(m, Pelem)(a, c) ∧ Pelem(c, b)), n = 0

Rule argument

spec: nat inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n, m] flags:

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m