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Proof step 22 from proof datasize-further (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
⊦ datasize(buf, n + 1) = (buf[n] = empty ⊃ datasize(buf, n);n + 1)
Proof step 24:
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[n] = empty ⊃ datasize(buf, n);n + 1) = datasize(buf, n1) ↔ (∀ n0. n < n0 ∧ n0 < n1 → 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 + 1 ≤ 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 +1 = m + 1; (s)