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Proof step 20 from proof datasize-further (previous step)

∀ n0. ¬ n < n0 → buf[n0] = empty, ∀ buf. 0 = datasize(buf, n) ↔ (∀ n0. n0 < n → buf[n0] = empty) ⊦ 0 = n + 1, buf[n] = empty

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
∀ buf. 0 = datasize(buf, n) ↔ (∀ n0. n0 < n → buf[n0] = empty), ∀ n0. ¬ n < n0 → buf[n0] = empty ⊦ buf[n] = 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)