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Proof step 18 from proof datasize-zero-h (previous step)

∀ n. ¬ m0 < n → buf[n] = empty, ∀ buf. datasize(buf, m0) = 0 ↔ (∀ n. n < m0 → buf[n] = empty) ⊦ (buf[m0] = empty ⊃ datasize(buf, m0);m0 + 1) = 0, buf[m0] = empty

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
∀ buf. datasize(buf, m0) = 0 ↔ (∀ n. n < m0 → buf[n] = empty), ∀ n. ¬ m0 < n → buf[n] = empty ⊦ buf[m0] = empty

Rule argument

Additional information

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