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Proof step 55 from proof datasize-further (previous step)

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

User applied rule structural induction

And obtained the following new goals

Proof step 56:
⊦ datasize(buf, 0) = 0
Proof step 57:
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

Rule argument

indvar: n1

Additional information

Used simplifier rules: datasize-base: ⊦ datasize(buf, 0) = 0;%nat freely generated by 0, +1;;% ⊦ ¬ m < 0; (s)% ⊦ m + 1 ≠ 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic