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Proof step 10 from proof datasize-further (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 11:
0 = (buf[n] = empty ⊃ datasize(buf, n);n + 1), ∀ buf. 0 = datasize(buf, n) ↔ (∀ n0. n0 < n → buf[n0] = empty) ⊦ ∀ n0. ¬ n < n0 → buf[n0] = empty
Proof step 16:
∀ buf. 0 = datasize(buf, n) ↔ (∀ n0. n0 < n → buf[n0] = empty), ∀ n0. ¬ n < n0 → buf[n0] = empty ⊦ 0 = (buf[n] = empty ⊃ datasize(buf, n);n + 1)

Rule argument

pos: right 1

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