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Proof step 27 from proof datasize-further (previous step)

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

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 28:
buf[n] = empty, datasize(buf, n) = datasize(buf, n1), n < n1, n < # buf, n0 < n1, n < n0, n + 1 < n1, n ≤ n1 ∧ n ≤ # buf → (datasize(buf, n) = datasize(buf, n1) ↔ (∀ n2. n ≤ n2 ∧ n2 < n1 → buf[n2] = empty)), ∀ n1, buf. n ≤ n1 ∧ n ≤ # buf → (datasize(buf, n) = datasize(buf, n1) ↔ (∀ n2. n ≤ n2 ∧ n2 < n1 → buf[n2] = empty)) ⊦ buf[n0] = empty

Rule argument

pos: left 8 terms: n1, buf

Additional information

Used terms: n1, buf