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Proof step 38 from proof datasize-further (previous step)

n + 1 = datasize(buf, n2), datasize(buf, n) = n + 1 ↔ (∀ n1. n ≤ n1 ∧ ¬ n2 < n1 → buf[n1] = empty), buf[n2] = empty, n < n2, n < n0, n < # buf, n0 ≤ n2, ∀ n, buf, n0. n + 1 = datasize(buf, n2) ∧ (datasize(buf, n) = datasize(buf, n2) ↔ (∀ n3. n ≤ n3 ∧ n3 < n2 → buf[n3] = empty)) ∧ n < n0 ∧ n0 < n2 ∧ n < # buf ∧ n < n2 ∧ datasize(buf, n2) < n2 ∧ datasize(buf, n2) ≤ n0 ∧ datasize(buf, n2) ≤ # buf → buf[n0] = empty ∨ buf[n] = empty ⊦ buf[n0] = empty, buf[n] = empty

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 39:
n + 1 = datasize(buf, n2), datasize(buf, n) = datasize(buf, n2) ↔ (∀ n1. n ≤ n1 ∧ ¬ n2 < n1 → buf[n1] = empty), buf[n2] = empty, n < # buf, n < n0, n < n2, n0 ≤ n2, datasize(buf, n2) ≤ # buf, datasize(buf, n2) ≤ n0, datasize(buf, n2) ≤ n2, ∀ n, buf, n0. n + 1 = datasize(buf, n2) ∧ (datasize(buf, n) = datasize(buf, n2) ↔ (∀ n3. n ≤ n3 ∧ n3 < n2 → buf[n3] = empty)) ∧ n < n0 ∧ n0 < n2 ∧ n < # buf ∧ n < n2 ∧ datasize(buf, n2) < n2 ∧ datasize(buf, n2) ≤ n0 ∧ datasize(buf, n2) ≤ # buf → buf[n0] = empty ∨ buf[n] = empty ⊦ buf[n] = empty, buf[n0] = empty

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat