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Proof step 165 from proof refine-read-buf (previous step)

# avol0 = # avol1, # lpt0 = # avol1, # buf2 = EB_PAGE_SIZE, min(n1, n2) + n4 = n1 + n5, n3 + n5 + n1 = # buf3, LEB_SIZE = n0 + # buf0 + # buf, avol0[m2] = mapped(buf0 + buf), avol1[m2] = mapped(buf0 + buf), n = n6 + n3, subarray(buf2, 0, n2) = subarray(buf, n3, n2), n3 < # buf3, n < # buf3, n3 + n2 < # buf, n3 < # buf, m2 < # avol1, n2 < EB_PAGE_SIZE, # buf0 < LEB_SIZE, (n3 + # buf0) + n2 < LEB_SIZE, n3 < n0 + # buf, awbuf-inv(avol1, buffered(m2)), ebm-vol-inv(avol0), is-aligned(n3 + # buf0, EB_PAGE_SIZE), n3 ≤ n, min(n1, n2) ≤ EB_PAGE_SIZE, min(n1, n2) ≤ n1 + n5, n1 + n3 ≤ # buf + n0 ⊦ n5 + n1 = 0, (subarray(buf2, 0, min(n1, n2)) + subarray(subarray(buf + mkempbuf(n0), 0, n1 + n3) + subarray(buf3, n1 + n3, n5), n3 + min(n1, n2), n4))[n6] = (n < n1 + n3 ⊃ (buf + mkempbuf(n0))[n];subarray(buf3, n1 + n3, n5)[n - n1 + n3])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 166:
# buf2 = EB_PAGE_SIZE, # lpt0 = # avol1, # avol0 = # avol1, n3 + n5 + n1 = # buf3, min(n1, n2) + n4 = n1 + n5, LEB_SIZE = n0 + # buf0 + # buf, avol1[m2] = mapped(buf0 + buf), avol0[m2] = mapped(buf0 + buf), subarray(buf2, 0, n2) = subarray(buf, n3, n2), n6 < n1 + n5, n6 < n5 + n1, n3 < n0 + # buf, (n3 + # buf0) + n2 < LEB_SIZE, # buf0 < LEB_SIZE, n2 < EB_PAGE_SIZE, m2 < # avol1, n3 < # buf, n3 + n2 < # buf, n3 < # buf3, awbuf-inv(avol1, buffered(m2)), ebm-vol-inv(avol0), is-aligned(n3 + # buf0, EB_PAGE_SIZE), n1 + n3 ≤ # buf + n0, min(n1, n2) ≤ n1 + n5, min(n1, n2) ≤ EB_PAGE_SIZE ⊦ (subarray(buf2, 0, min(n1, n2)) + subarray(subarray(buf + mkempbuf(n0), 0, n1 + n3) + subarray(buf3, n1 + n3, n5), n3 + min(n1, n2), n4))[n6] = (n6 < n1 ⊃ (buf + mkempbuf(n0))[n6 + n3];subarray(buf3, n1 + n3, n5)[n6 - n1])

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m = m0 + n0 ∧ n < n0 → m0 + n < m; (s)% ⊦ m < n → n ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ m ≤ m + n; (s)% ⊦ m + n - n + n1 = m - n1; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n < m + n0 ↔ n < n0; (s)