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Proof step 73 from proof datasize-further (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: n0

Additional information

Used simplifier rules: ⊦ ¬ n < n; (s) Used terms: n0