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Proof step 5 from proof datasize-subarray0-h (previous step)

m1 +1 ≤ # buf, m ≤ # buf, m1 +1 ≤ m, ∀ m, buf. m1 ≤ m ∧ m ≤ # buf ∧ m1 ≤ # buf → datasize(subarray(buf, 0, m), m1) = datasize(buf, m1) ⊦ datasize(subarray(buf, 0, m), m1 +1) = datasize(buf, m1 +1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ datasize(buf, n + 1) = (buf[n] = empty ⊃ datasize(buf, n);n + 1)
Proof step 7:
m1 < # buf, m1 < m, m ≤ # buf, ∀ m, buf. m1 ≤ m ∧ m ≤ # buf ∧ m1 ≤ # buf → datasize(subarray(buf, 0, m), m1) = datasize(buf, m1) ⊦ (buf[m1] = empty ⊃ datasize(subarray(buf, 0, m), m1);m1 + 1) = (buf[m1] = empty ⊃ datasize(buf, m1);m1 + 1)

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)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ n + 0 = n; (s)% ⊦ n0 < n → subarray(buf, m, n)[n0] = buf[m + n0]; (s)% ⊦ 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)