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Proof step 19 from proof modify+-sym (previous step)

modify(rs)(H0, H), rtc(n0, modify(rs))(H, H1), ∀ rs, H0, H1. rtc(n0, modify(rs))(H0, H1) → rtc(n0, modify(rs))(H1, H0) ⊦ n0 + 1 = 0, ∃ H2. rtc(n0 + 1 - 1, modify(rs))(H1, H2) ∧ modify(rs)(H2, H0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
modify(rs)(H0, H), rtc(n0, modify(rs))(H, H1), ∀ rs, H0, H1. rtc(n0, modify(rs))(H0, H1) → rtc(n0, modify(rs))(H1, H0) ⊦ ∃ H2. rtc(n0, modify(rs))(H1, H2) ∧ modify(rs)(H2, H0)

Rule argument

Additional information

Used simplifier rules: ⊦ n + 0 = n; (s)% ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)