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Proof step 14 from proof datasize-subarray-h (previous step)

m0 + m1 = # buf, buf[m0 + m2] = empty, datasize(buf, m0 + m2) = n0 + m0, m2 < m1, m0 < datasize(buf, m0 + m2), m0 ≤ datasize(buf, m0 + m2), ∀ 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) = n0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
m0 + m1 = # buf, buf[m0 + m2] = empty, datasize(buf, m0 + m2) = n0 + m0, 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) ⊦ datasize(subarray(buf, m0, m1), m2) = n0, n0 = 0

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < m + n ↔ n ≠ 0; (s)% ⊦ m < n → n ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n → m ≤ n; (s)