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Proof step 72 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, ∀ 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 ⊦ n + 1 = n0 + 1, buf[n] = empty, buf[n0] = empty

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 73:
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

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)