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Proof step 2 from proof le-mapped (previous step)

avol0[n].mapped?, n < # avol0, avol0 ⊆ avol1 ⊦ avol1[n].mapped?

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
# avol0 = # avol1, avol1[n] = unmapped, avol0[n] = mapped(buf), n < # avol1, ∀ m. m < # avol1 → (avol0[m].mapped? → avol1[m] = avol0[m]) ∧ (avol0[m] = unmapped ∧ avol1[m].mapped? → # avol1[m].data ≤ LEB_SIZE ∧ is-aligned(# avol1[m].data, EB_PAGE_SIZE)) ⊦

Rule argument

Additional information

Used simplifier rules: ebm-vol-le: ⊦ avol0 ⊆ avol1 ↔ # avol0 = # avol1 ∧ (∀ m. m < # avol0 → (avol0[m].mapped? → avol1[m] = avol0[m]) ∧ ( avol0[m] = unmapped ∧ avol1[m].mapped? → # avol1[m].data ≤ LEB_SIZE ∧ is-aligned(# avol1[m].data, EB_PAGE_SIZE)));% ⊦ m < n → n ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ aleb.mapped? ↔ aleb ≠ unmapped; (ss)% ⊦ aleb.mapped? ↔ (∃ buf. aleb = mapped(buf)); (as)