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Proof step 15 from proof rtc-last (previous step)

∀ Pelem, a, c, b. rtc(m0, Pelem)(a, c) ∧ Pelem(c, b) → (∃ c. Pelem(a, c) ∧ rtc(m0, Pelem)(c, b)), rtc(m0 +1, Pelem)(a, c), Pelem(c, b) ⊦ ∃ c. Pelem(a, c) ∧ rtc(m0 +1, Pelem)(c, b)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
Pelem(c, b), Pelem(a, c0), rtc(m0, Pelem)(c0, c), ∀ Pelem, a, c, b. rtc(m0, Pelem)(a, c) ∧ Pelem(c, b) → (∃ c. Pelem(a, c) ∧ rtc(m0, Pelem)(c, b)) ⊦ ∃ c1. Pelem(a, c1) ∧ (∃ c. Pelem(c1, c) ∧ rtc(m0, Pelem)(c, b))

Rule argument

Additional information

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