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Proof step 149 from proof refine-read-buf (previous step)

# avol0 = # avol1, # lpt0 = # avol1, # buf2 = EB_PAGE_SIZE, n1 + n5 = n4 + min(n1, n2), n3 + n5 + n1 = # buf3, LEB_SIZE = n0 + # buf4, avol0[m2] = mapped(buf4), avol1[m2] = mapped(buf4), subarray(buf2, 0, n2) = subarray(buf4, n3 + m0, n2), (n3 + m0) + n2 < # buf4, n3 + m0 < # buf4, m2 < # avol1, n2 < EB_PAGE_SIZE, m0 < LEB_SIZE, (n3 + m0) + n2 < LEB_SIZE, n3 + m0 < LEB_SIZE, awbuf-inv(avol1, buffered(m2)), ebm-vol-inv(avol0), is-aligned(n3 + m0, EB_PAGE_SIZE), # buf4 ≤ LEB_SIZE, min(n1, n2) ≤ n1 + n5, n1 + n3 + m0 ≤ LEB_SIZE, min(n1, n2) ≤ EB_PAGE_SIZE, n3 ≤ # buf3, ∃ m1. m0 + m1 = # buf4 ∧ m1 ≠ 0 ⊦ subarray(buf4, m0, n3) + subarray(buf2, 0, min(n1, n2)) + subarray(subarray(buf4 + mkempbuf(n0), m0, n1 + n3) + subarray(buf3, n1 + n3, n5), n3 + min(n1, n2), n4) = subarray(buf4 + mkempbuf(n0), m0, n1 + n3) + subarray(buf3, n1 + n3, n5)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 150:
# buf2 = EB_PAGE_SIZE, # lpt0 = # avol1, # avol0 = # avol1, m0 + m = # buf4, n3 + n5 + n1 = # buf3, n1 + n5 = n4 + min(n1, n2), LEB_SIZE = n0 + m0 + m, avol1[m2] = mapped(buf4), avol0[m2] = mapped(buf4), subarray(buf2, 0, n2) = subarray(buf4, n3 + m0, n2), n3 + m0 < LEB_SIZE, (n3 + m0) + n2 < LEB_SIZE, m0 < LEB_SIZE, n2 < EB_PAGE_SIZE, m2 < # avol1, n3 < m, n3 + n2 < m, awbuf-inv(avol1, buffered(m2)), ebm-vol-inv(avol0), is-aligned(n3 + m0, EB_PAGE_SIZE), n1 + n3 ≤ m + n0, min(n1, n2) ≤ n1 + n5, n3 ≤ # buf3, min(n1, n2) ≤ EB_PAGE_SIZE ⊦ n0 + m = 0, subarray(buf4, m0, n3) + subarray(buf2, 0, min(n1, n2)) + subarray(subarray(buf4 + mkempbuf(n0), m0, n1 + n3) + subarray(buf3, n1 + n3, n5), n3 + min(n1, n2), n4) = subarray(buf4 + mkempbuf(n0), m0, n1 + n3) + subarray(buf3, n1 + n3, n5)

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)% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ m + n < m0 → m + n < m0 + n0; (s)% ⊦ m < m + n ↔ n ≠ 0; (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