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Proof step 59 from proof datasize-further (previous step)

datasize(buf, n) = (buf[n0] = empty ⊃ datasize(buf, n0);n0 + 1) ↔ (∀ n1. n ≤ n1 ∧ ¬ n0 < n1 → buf[n1] = empty), n < # buf, n ≤ n0, ∀ buf, n. (datasize(buf, n) = datasize(buf, n0) ↔ (∀ n2. n ≤ n2 ∧ n2 < n0 → buf[n2] = empty)) ∧ n < n0 ∧ n < # buf ∧ (∀ n2. n < n2 ∧ n2 < n0 → buf[n2] = empty) → n + 1 = datasize(buf, n0) ∨ buf[n] = empty, ∀ n1. n < n1 ∧ ¬ n0 < n1 → buf[n1] = empty ⊦ n + 1 = (buf[n0] = empty ⊃ datasize(buf, n0);n0 + 1), buf[n] = empty

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

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