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Proof step 35 from proof datasize-further (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 36:
⊦ datasize(buf, n + 1) = (buf[n] = empty ⊃ datasize(buf, n);n + 1)
Proof step 37:
n + 1 = (buf[n2] = empty ⊃ datasize(buf, n2);n2 + 1), datasize(buf, n) = n + 1 ↔ (∀ n1. n ≤ n1 ∧ ¬ n2 < n1 → buf[n1] = empty), n < # buf, n < n0, n < n2, 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[n] = empty, buf[n0] = empty

Rule argument

Additional information

Used simplifier rules: datasize-rec: ⊦ datasize(buf, n + 1) = (buf[n] = empty ⊃ datasize(buf, n);n + 1);% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n → m ≤ n; (s)% ⊦ m + 1 ≤ n ↔ m < n; (s)%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n + 1 ↔ ¬ n < m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m +1 = m + 1; (s)