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Proof step 2 from proof mapped-full-range-eq (previous step)

mapped-full?(avol1, n0, n1), range=(avol0, m, avol1, m, m2), n0 + n1 ≤ m + m2, m ≤ n0 ⊦ mapped-full?(avol0, n0, n1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
m < n0 + n1, n0 < # avol0, n < m + m2, n < n0 + n1, n < # avol0, m < # avol0, n0 < m + m2, range=(avol0, m, avol1, m, m2), m ≤ n0, n0 + n1 ≤ m + m2, n0 ≤ n, m ≤ n, ∀ n2. n0 ≤ n2 ∧ n2 < n0 + n1 ∧ n2 < # avol1 → avol1[n2].mapped? ∧ # avol1[n2].data = LEB_SIZE ⊦ m2 = 0, n1 = 0, avol0[n].mapped? ∧ # avol0[n].data = LEB_SIZE

Rule argument

Additional information

Used simplifier rules: mapped-full: ⊦ mapped-full?(avol, n, m) ↔ (∀ n0. n ≤ n0 ∧ n0 < n + m ∧ n0 < # avol → avol[n0].mapped? ∧ # avol[n0].data = LEB_SIZE);%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)%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic