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Proof step 15 from proof refine-read-buf (previous step)

# avol0 = # avol1, # lpt0 = # avol1, # buf2 = EB_PAGE_SIZE, avol1[m2] = mapped(buf1), m2 < # avol1, abs(avol1, bleb, mkwbuf(buf2, m6, m4), avol0, rofs), awbuf-inv(avol1, bleb), ebm-vol-inv(avol0), n ≤ # buf3, m0 + n ≤ LEB_SIZE, bleb ≠ nobuffer → bleb.leb < # avol1 ∧ avol0[bleb.leb].mapped? ∧ m4 < EB_PAGE_SIZE ∧ (¬ rofs → # avol0[bleb.leb].data = m6) ∧ (rofs → m6 ≤ # avol0[bleb.leb].data) ∧ is-aligned(m6, EB_PAGE_SIZE) ⊦ ⟪choose err0 with ⌊err0⌋ ∨ err0 = ESUCCESS in err := err0; buf0 := buf3; if err = ESUCCESS then buf0 := copy(avol0[m2].data + mkempbuf(LEB_SIZE - # avol0[m2].data), m0, buf0, 0, n) else {n0 := 0; choose buf with # buf = # buf0 ∧ range=(buf0, 0, buf, 0, n0) ∧ range=(buf0, n0 + n, buf, n0 + n, # buf0 - n0 + n) in buf0 := buf}; buf := buf0; if err = ESUCCESS ∧ buffered(m2) = bleb ∧ m0 ≤ m6 + m4 ∧ m6 ≤ m0 + n then let m1 = (m0 ≤ m6 ⊃ 0;m0 - m6), m = (m0 ≤ m6 ⊃ m6 - m0;0) in buf := copy(buf2, m1, buf, m, min(n - m, m4 - m1))⟫ ⟨awbuf_read_buf#(m2, m0, n, adr0, m5, plog0, avol1, lpt0, ns0, bleb, rofs; buf3; err0)⟩ (buf = buf3 ∧ err = err0)

Heuristic symbolic execution applied rule choose right

And obtained the following new goals

Rule argument

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