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Proof step 19 from proof rtc+-unfold (previous step)

p(a, c), rtc(n, c, b) ⊦ n + 1 ≠ 0 ∧ rtc(n + 1, a, b), ∃ n. n ≠ 0 ∧ rtc(n, a, b)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
p(a, c), rtc(n, c, b) ⊦ ∃ n. n ≠ 0 ∧ rtc(n, a, b), ∃ c. p(a, c) ∧ rtc(n, c, b)

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%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2