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Proof step 6 from proof datasize-subarray-h (previous step)

m0 + m1 = # buf, m2 +1 ≤ m1, ∀ m0, m1, buf. m0 + m1 = # buf ∧ m2 ≤ m1 → datasize(subarray(buf, m0, m1), m2) = (datasize(buf, m0 + m2) ≤ m0 ⊃ 0;datasize(buf, m0 + m2) - m0) ⊦ datasize(subarray(buf, m0, m1), m2 +1) = (datasize(buf, m0 + m2 +1) ≤ m0 ⊃ 0;datasize(buf, m0 + m2 +1) - m0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
⊦ datasize(buf, n + 1) = (buf[n] = empty ⊃ datasize(buf, n);n + 1)
Proof step 8:
m0 + m1 = # buf, m2 < m1, ∀ m0, m1, buf. m0 + m1 = # buf ∧ m2 ≤ m1 → datasize(subarray(buf, m0, m1), m2) = (datasize(buf, m0 + m2) ≤ m0 ⊃ 0;datasize(buf, m0 + m2) - m0) ⊦ (buf[m0 + m2] = empty ⊃ datasize(subarray(buf, m0, m1), m2);m2 + 1) = ((buf[m0 + m2] = empty ⊃ datasize(buf, m0 + m2);(m0 + m2) + 1) ≤ m0 ⊃ 0;(buf[m0 + m2] = empty ⊃ datasize(buf, m0 + m2);(m0 + m2) + 1) - m0)

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% ⊦ 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)