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Proof step 10 from proof lebs-range-eq (previous step)

n0 + n < m + m2, n0 + n < # avol0, range=(avol0, m, avol1, m, m2), m ≤ n0, ∀ avol0, m, avol1, m2, n0. range=(avol0, m, avol1, m, m2) ∧ n0 + n ≤ m + m2 ∧ m ≤ n0 → lebs(avol0, n0, n) = lebs(avol1, n0, n) ⊦ lebs(avol0, n0, n) + (avol0[n0 + n].mapped? ⊃ avol0[n0 + n].data + mkempbuf(LEB_SIZE - # avol0[n0 + n].data);mkempbuf(LEB_SIZE)) = lebs(avol1, n0, n + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
n0 + n < m + m2, n0 + n < # avol0, range=(avol0, m, avol1, m, m2), m ≤ n0, ∀ avol0, m, avol1, m2, n0. range=(avol0, m, avol1, m, m2) ∧ n0 + n ≤ m + m2 ∧ m ≤ n0 → lebs(avol0, n0, n) = lebs(avol1, n0, n) ⊦ lebs(avol0, n0, n) + (avol0[n0 + n].mapped? ⊃ avol0[n0 + n].data + mkempbuf(LEB_SIZE - # avol0[n0 + n].data);mkempbuf(LEB_SIZE)) = lebs(avol1, n0, n + 1)

Rule argument

Additional information

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