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Proof step 225 from proof refine-write-buf (previous step)

# avol0 = # avol1, # lpt0 = # avol1, # buf1 = EB_PAGE_SIZE, avol0[m4] = mapped(buf2), align↓(m2 + n3, EB_PAGE_SIZE) = n + m2, m2 < EB_PAGE_SIZE, m4 < # avol1, abs(avol1, buffered(m4), mkwbuf(buf1, # buf2, m2), avol0, false), awbuf-inv(avol1, buffered(m4)), ebm-vol-inv(avol0), is-aligned(n + m2, EB_PAGE_SIZE), is-aligned(# buf2, EB_PAGE_SIZE), # avol1[m4].data + n3 ≤ LEB_SIZE, n3 ≤ # buf3, EB_PAGE_SIZE ≤ m2 + n3, # buf2 + m2 + n3 ≤ LEB_SIZE, n ≤ # buf3, # buf2 + n + m2 ≤ LEB_SIZE ⊦ n3 = 0, ∃ err0. ⌊err0⌋ ∨ err0 = ESUCCESS

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 2 terms: ECOMMIT

Additional information

Used simplifier rules: ⊦ EB_PAGE_SIZE ≠ 0; (s)% ⊦ ESUCCESS ≠ ECOMMIT; (s)% ⊦ ⌊ECOMMIT⌋ ; (s)%n ≠ 0, m % n = 0 ⊦ align↓(m, n) = m; (s)% ⊦ m + n = m + n0 ↔ n = n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%m ≠ 0, is-aligned(n1, m) ⊦ is-aligned(n0 + n1, m) ↔ is-aligned(n0, m); (s) Used terms: ECOMMIT