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Proof step 30 from proof datasize-further (previous step)

n + 1 = 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, buf[n] = empty

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 31:
n + 1 = datasize(buf, n1), n < n0, n0 < n1, n < # buf, n < n1, datasize(buf, n1) < n1, datasize(buf, n1) ≤ n0, datasize(buf, n1) ≤ # buf, ∀ n1, buf. n ≤ n1 ∧ n ≤ # buf → (datasize(buf, n) = datasize(buf, n1) ↔ (∀ n2. n ≤ n2 ∧ n2 < n1 → buf[n2] = empty)) ⊦ buf[n0] = empty, buf[n] = empty

Rule argument

Additional information

Used simplifier rules: fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n → m ≤ n; (s)%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%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