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Proof step 2 from proof mapped-full-aligndown-vol (previous step)

mapped-full?(avol0, n, m), avol0 ⊆ avol1 ⊦ mapped-full?(avol1, n, m)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n0 < # avol1, n0 < n + m, n < # avol1, n ≤ n0, avol0 ⊆ avol1, ∀ n0. n ≤ n0 ∧ n0 < n + m ∧ n0 < # avol0 → avol0[n0].mapped? ∧ # avol0[n0].data = LEB_SIZE ⊦ m = 0, avol1[n0].mapped? ∧ # avol1[n0].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)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic