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Proof step 85 from proof refine-write-buf (previous step)

# buf1 = EB_PAGE_SIZE, # lpt0 = # avol1, # avol0 = # avol1, avol0[m4] = mapped(buf2), align↓(m2 + n3, EB_PAGE_SIZE) = n + m2, n3 < n + (n3 - n), # buf3 < n + (n3 - n), m2 < EB_PAGE_SIZE, m4 < # avol1, abs(avol1, buffered(m4), mkwbuf(buf1, # buf2, m2), avol0, false), awbuf-inv(avol1, buffered(m4)), ebm-vol-inv(avol0), is-aligned(# buf2, EB_PAGE_SIZE), is-aligned(n + m2, EB_PAGE_SIZE), # buf2 + n + m2 ≤ LEB_SIZE, # buf2 + m2 + n3 ≤ LEB_SIZE, EB_PAGE_SIZE ≤ m2 + n3, n3 ≤ # buf3, # avol1[m4].data + n3 ≤ LEB_SIZE, n ≤ # buf3 ⊦ n + m2 = 0, n3 - n = 0, n3 = 0

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [n, m, n0] → [n, n3, n0]

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic% ⊦ EB_PAGE_SIZE ≠ 0; (s)%n ≠ 0 ⊦ is-aligned(n0, n) → align↓(n0 + n1, n) = n0 + align↓(n1, n); (s)% ⊦ m < n → n ≠ 0; (s)% ⊦ m ≤ m + n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n → (n ≤ m ↔ false); (s)%n ≠ 0, m % n = 0 ⊦ align↓(m, n) = m; (s)% ⊦ m + n = m + n0 ↔ n = n0; (s)% ⊦ m = m + n ↔ n = 0; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)%m ≠ 0 ⊦ align↓(n0 + n1, m) = n2 + n1 → ¬ n0 < n2; (s)%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n, n3, n0