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Proof step 41 from proof datasize-further (previous step)

n + 1 = datasize(buf, n2), datasize(buf, n) = datasize(buf, n2) ↔ (∀ n1. n ≤ n1 ∧ ¬ n2 < n1 → buf[n1] = empty), buf[n2] = empty, n < # buf, n < n0, n < n2, n0 ≤ n2, datasize(buf, n2) ≤ # buf, datasize(buf, n2) ≤ n0, datasize(buf, n2) ≤ n2 ⊦ buf[n] = empty, buf[n0] = empty, (datasize(buf, n) = datasize(buf, n2) ↔ (∀ n3. n ≤ n3 ∧ n3 < n2 → buf[n3] = empty)) ∧ n0 < n2 ∧ datasize(buf, n2) < n2

User applied rule case distinction

And obtained the following new goals

Proof step 42:
n + 1 = datasize(buf, n2), datasize(buf, n) = datasize(buf, n2) ↔ (∀ n1. n ≤ n1 ∧ ¬ n2 < n1 → buf[n1] = empty), buf[n2] = empty, n < n2, n < n0, n < # buf, datasize(buf, n2) ≤ n2, datasize(buf, n2) ≤ n0, datasize(buf, n2) ≤ # buf, n0 ≤ n2 ⊦ datasize(buf, n) = datasize(buf, n2) ↔ (∀ n3. n ≤ n3 ∧ n3 < n2 → buf[n3] = empty), buf[n0] = empty, buf[n] = empty
Proof step 47:
⊦ datasize(buf, n + 1) = (buf[n] = empty ⊃ datasize(buf, n);n + 1)

Rule argument

pos: right 3

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 → (n ≤ m ↔ m = n); (s)%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat% ⊦ m + n = m + n0 ↔ n = n0; (s)% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ ¬ n < n; (s)% ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1); (s)