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Proof step 279 from proof refine-read-buf (previous step)

# lpt0 = # avol1, # avol0 = # avol1, # buf3 = n4 + n2, n1 + n5 = n4 + n2, (# buf + # buf4) + n3 + n2 = LEB_SIZE, # buf + # buf0 = EB_PAGE_SIZE, avol1[m2] = mapped(buf4 + buf + subarray(buf0, 0, n2)), avol0[m2] = mapped(buf4), # buf < EB_PAGE_SIZE, m2 < # avol1, n2 < # buf0, awbuf-inv(avol1, buffered(m2)), ebm-vol-inv(avol0), is-aligned(# buf4, EB_PAGE_SIZE), # buf4 ≤ LEB_SIZE, n2 ≤ n1, n1 ≤ n2 + n3 ⊦ # buf = 0, EB_PAGE_SIZE = 1, subarray(mkempbuf(n1) + subarray(buf3, n1, n5), n2, n4) = subarray(mkempbuf(n3), 0, n1 - n2) + subarray(buf3, n1, n5)

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] → [n2, n1, n]

Additional information

Used simplifier rules: ⊦ m + n = m + n0 ↔ n = n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ # subarray(buf, m, n) = n; (s)% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ ¬ n < n; (s)%n2 ≤ n1 + # buf ⊦ subarray(mkempbuf(n0 + n1) + buf, n0, n2) = subarray(mkempbuf(n1) + buf, 0, n2); (s)% ⊦ #(buf + buf0) = # buf + # buf0; (s)% ⊦ # mkempbuf(n) = n; (s)%append-a: ⊦ (buf + buf0) + buf1 = buf + buf0 + buf1; from specification array-splice[buffer]%# buf = n ⊦ subarray(buf, 0, n) = buf; (s)% ⊦ buf + buf0 = buf1 + buf0 ↔ buf = buf1; (s)% ⊦ m ≤ m + n; (s)% ⊦ n1 ≤ n0 → subarray(mkempbuf(n0), 0, n1) = mkempbuf(n1); (s)% ⊦ mkempbuf(n) = mkempbuf(m) ↔ n = m; (s)%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n2, n1, n