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Proof step 14 from proof rtc-last (previous step)

rtc(m, Pelem)(a, c), Pelem(c, b) ⊦ ∃ c. Pelem(a, c) ∧ rtc(m, Pelem)(c, b)

User applied rule structural induction

And obtained the following new goals

Proof step 15:
∀ 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)

Rule argument

indvar: m

Additional information

Used simplifier rules: rtc-base: ⊦ rtc(0, Pelem)(a, b) ↔ a = b;%nat freely generated by 0, +1;;