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Proof step 2 from proof le-size-0 (previous step)

# avol0 < # avol1, avol0 ⊆ avol1 ⊦

Heuristic simplifier applied rule simplifier

And closed the goal!

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 < m; (s)% ⊦ m < n → n ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic