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Proof step 16 from proof refine-write-buf (previous step)

# buf1 = EB_PAGE_SIZE, # lpt0 = # avol1, # avol0 = # avol1, rofs ↔ rofs0, avol0[m4] = mapped(buf2), adr = adr0, avol = avol1, bleb = buffered(m4), lpt = lpt0, m0 = m3, n0 = n, ns = ns0, plog = plog0, rofs0, m2 < EB_PAGE_SIZE, m4 < # avol1, abs(avol1, buffered(m4), mkwbuf(buf1, m1, m2), avol0, true), awbuf-inv(avol1, buffered(m4)), ebm-vol-inv(avol0), is-aligned(m1, EB_PAGE_SIZE), n ≤ # buf, # avol1[m4].data + n ≤ LEB_SIZE, m1 ≤ # buf2 ⊦ ⟨if bleb ≠ nobuffer ∧ # avol[bleb.leb].data + n0 ≤ LEB_SIZE ∧ n0 ≤ # buf then if rofs then err := EROFS else {fail_or_success#(; ; err); let m = bleb.leb in if err = ESUCCESS then avol[m] := mapped(avol[m].data + subarray(buf, 0, n0)) else { choose n with n ≠ 0 ∧ n < n0 ∧ is-aligned(# avol[m].data + n, EB_PAGE_SIZE) in avol[m] := mapped(avol[m].data + subarray(buf, 0, n)) ifnone skip or skip; rofs := true}} else abort⟩ (abs(avol, buffered(m4), mkwbuf(buf1, m1, m2), avol0, true) ∧ rofs ∧ EROFS = err)

Heuristic symbolic execution applied rule if positive right

And obtained the following new goals

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ buffered(m).leb = m; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n → (n ≤ m ↔ false); (s)%n < # avol ⊦ abs(avol, bleb, wbuf, avol0, rofs) ∧ avol0[n] = mapped(buf) → (avol[n].mapped? ↔ true); (ss)% ⊦ buffered(m).isbuf?; (s)% ⊦ nobuffer ≠ buffered(m); (s)