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Proof step 18 from proof datasize-further (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
buf[n] = empty, n0 < n, ∀ buf. 0 = datasize(buf, n) ↔ (∀ n0. n0 < n → buf[n0] = empty), ∀ n0. ¬ n < n0 → buf[n0] = empty ⊦ 0 = datasize(buf, n), buf[n0] = empty

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic