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Proof step 225 from proof refine-read-buf (previous step)

# buf3 = n5 + n1, # avol0 = # avol1, # lpt0 = # avol1, # buf2 = EB_PAGE_SIZE, n1 + n5 = n4 + min(n1, n2), LEB_SIZE = n0 + # buf0 + # buf, avol0[m2] = mapped(buf0 + buf), avol1[m2] = mapped(buf0 + buf), subarray(buf2, 0, n2 + n3) = subarray(buf, 0, n2 + n3), n < n1 + n5, n3 < n0 + # buf, (n3 + # buf0) + n2 < LEB_SIZE, n3 < EB_PAGE_SIZE, m2 < # avol1, n2 + n3 < EB_PAGE_SIZE, n3 < # buf, # buf0 < LEB_SIZE, # buf0 + 1 < LEB_SIZE, n3 + n2 < # buf, n < n5 + n1, awbuf-inv(avol1, buffered(m2)), ebm-vol-inv(avol0), is-aligned(# buf0, EB_PAGE_SIZE), n3 + n1 ≤ # buf + n0, n3 + min(n1, n2) ≤ EB_PAGE_SIZE, min(n1, n2) ≤ n1 + n5 ⊦ # buf = 1, EB_PAGE_SIZE = 1, subarray(subarray(buf + mkempbuf(n0), n3, n1) + subarray(buf3, n1, n5), min(n1, n2), n4)[n - min(n1, n2)] = (n < n1 ⊃ (buf + mkempbuf(n0))[n3 + n];subarray(buf3, n1, n5)[n - n1]), n3 = 0, n < min(n1, n2)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 226:
# buf2 = EB_PAGE_SIZE, # lpt0 = # avol1, # avol0 = # avol1, # buf3 = n5 + n1, n1 + n5 = n4 + min(n1, n2), LEB_SIZE = n0 + # buf0 + # buf, avol1[m2] = mapped(buf0 + buf), avol0[m2] = mapped(buf0 + buf), subarray(buf2, 0, n2 + n3) = subarray(buf, 0, n2 + n3), n < n5 + n1, n3 + n2 < # buf, # buf0 + 1 < LEB_SIZE, # buf0 < LEB_SIZE, n3 < # buf, n2 + n3 < EB_PAGE_SIZE, m2 < # avol1, n3 < EB_PAGE_SIZE, (n3 + # buf0) + n2 < LEB_SIZE, n3 < n0 + # buf, n < n1 + n5, min(n1, n2) < n1 + n5, min(n1, n2) < n5 + n1, awbuf-inv(avol1, buffered(m2)), ebm-vol-inv(avol0), is-aligned(# buf0, EB_PAGE_SIZE), n3 + min(n1, n2) ≤ EB_PAGE_SIZE, n3 + n1 ≤ # buf + n0, min(n1, n2) ≤ n ⊦ # buf = 1, EB_PAGE_SIZE = 1, subarray(subarray(buf + mkempbuf(n0), n3, n1) + subarray(buf3, n1, n5), min(n1, n2), n4)[n - min(n1, n2)] = (n < n1 ⊃ (buf + mkempbuf(n0))[n3 + n];subarray(buf3, n1, n5)[n - n1]), n3 = 0, n4 = 0

Rule argument

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < m + n ↔ n ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m < n → m ≤ n; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)