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Proof step 57 from proof datasize-further (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 58:
⊦ datasize(buf, n + 1) = (buf[n] = empty ⊃ datasize(buf, n);n + 1)
Proof step 59:
datasize(buf, n) = (buf[n0] = empty ⊃ datasize(buf, n0);n0 + 1) ↔ (∀ n1. n ≤ n1 ∧ ¬ n0 < n1 → buf[n1] = empty), n < # buf, n ≤ n0, ∀ 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, ∀ n1. n < n1 ∧ ¬ n0 < n1 → buf[n1] = empty ⊦ n + 1 = (buf[n0] = empty ⊃ datasize(buf, n0);n0 + 1), buf[n] = 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 ≤ 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)