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Proof step 201 from proof refine-read-buf (previous step)

# buf2 = EB_PAGE_SIZE, # lpt0 = # avol1, # avol0 = # avol1, # buf3 = n5 + n1, n1 + n5 = n4 + min(n1, n2), LEB_SIZE = n0 + # buf4, avol1[m2] = mapped(buf4), avol0[m2] = mapped(buf4), subarray(buf2, 0, n2 + n3) = subarray(buf4, m6, n2 + n3), (n3 + m6) + n2 < # buf4, m6 + 1 < LEB_SIZE, m6 + 1 < # buf4, m6 < LEB_SIZE, n3 + m6 < # buf4, n2 + n3 < EB_PAGE_SIZE, m2 < # avol1, n3 < EB_PAGE_SIZE, (n3 + m6) + n2 < LEB_SIZE, n3 + m6 < LEB_SIZE, awbuf-inv(avol1, buffered(m2)), ebm-vol-inv(avol0), is-aligned(m6, EB_PAGE_SIZE), min(n1, n2) ≤ n1 + n5, # buf4 ≤ LEB_SIZE, n3 + min(n1, n2) ≤ EB_PAGE_SIZE, (n3 + m6) + n1 ≤ LEB_SIZE, ∃ m0. m6 + m0 = # buf4 ∧ m0 ≠ 0 ⊦ subarray(buf2, n3, min(n1, n2)) + subarray(subarray(buf4 + mkempbuf(n0), n3 + m6, n1) + subarray(buf3, n1, n5), min(n1, n2), n4) = subarray(buf4 + mkempbuf(n0), n3 + m6, n1) + subarray(buf3, n1, n5), EB_PAGE_SIZE = 1, n3 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 202:
# buf3 = n5 + n1, # avol0 = # avol1, # lpt0 = # avol1, # buf2 = EB_PAGE_SIZE, m6 + m = # buf4, n1 + n5 = n4 + min(n1, n2), LEB_SIZE = n0 + m6 + m, avol0[m2] = mapped(buf4), avol1[m2] = mapped(buf4), subarray(buf2, 0, n2 + n3) = subarray(buf4, m6, n2 + n3), n3 + m6 < LEB_SIZE, (n3 + m6) + n2 < LEB_SIZE, n3 < EB_PAGE_SIZE, m2 < # avol1, n2 + n3 < EB_PAGE_SIZE, n3 < m, m6 < LEB_SIZE, m6 + 1 < LEB_SIZE, n3 + n2 < m, awbuf-inv(avol1, buffered(m2)), ebm-vol-inv(avol0), is-aligned(m6, EB_PAGE_SIZE), n3 + n1 ≤ m + n0, n3 + min(n1, n2) ≤ EB_PAGE_SIZE, min(n1, n2) ≤ n1 + n5 ⊦ n0 + m = 0, subarray(buf2, n3, min(n1, n2)) + subarray(subarray(buf4 + mkempbuf(n0), n3 + m6, n1) + subarray(buf3, n1, n5), min(n1, n2), n4) = subarray(buf4 + mkempbuf(n0), n3 + m6, n1) + subarray(buf3, n1, n5), EB_PAGE_SIZE = 1, m = 1, n3 = 0

Rule argument

Additional information

Used simplifier rules: nat0: ⊦ n0 = n1 + n → (n1 + m ≤ n0 ↔ m ≤ n);% ⊦ m = m0 + n0 ∧ n < n0 → m0 + n < m; (s)% ⊦ m ≤ m + n; (s)% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s)% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ m < m + n ↔ n ≠ 0; (s)% ⊦ m + n < m0 → m + n < m0 + n0; (s)% ⊦ 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%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic