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

p*(b, c), p(a, c0), rtc(n0, c0, b), rtc(n, c0, c) ⊦ rtc(n + 1, a, c), ∃ n. rtc(n, a, c)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
p*(b, c), p(a, c0), rtc(n0, c0, b), rtc(n, c0, c) ⊦ ∃ n. rtc(n, a, c), ∃ c0. p(a, c0) ∧ rtc(n, c0, c)

Rule argument

Additional information

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