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

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

User applied rule structural induction

And obtained the following new goals

Proof step 7:
∀ a, c, b. p(a, c) ∧ rtc(m0, c, b) → (∃ c. rtc(m0, a, c) ∧ p(c, b)), p(a, c), rtc(m0 +1, c, b) ⊦ ∃ c. rtc(m0 +1, a, c) ∧ p(c, b)

Rule argument

indvar: m

Additional information

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