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Proof step 12 from proof datasize-subarray-h (previous step)

m0 + m1 = # buf, buf[m0 + m2] = empty, 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) = (buf[m0 + m2] = empty ⊃ datasize(buf, m0 + m2);(m0 + m2) + 1) - m0, (buf[m0 + m2] = empty ⊃ datasize(buf, m0 + m2);(m0 + m2) + 1) ≤ m0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
m0 + m1 = # buf, buf[m0 + m2] = empty, m0 < datasize(buf, m0 + m2), 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) = datasize(buf, m0 + m2) - m0

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss)