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Proof step 13 from proof datasize-subarray-h (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 14:
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

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [datasize(buf, m0 + m2), m0, n]

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s)%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: datasize(buf, m0 + m2), m0, n