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Proof step 42 from proof valid-wl-write-unmapped (previous step)

# apebs = # wlar, avhdr1.sqn = valid-sqns(VTBL_VOLID, VTBL_LNUM, apebs).max, apebs[n] = mkpeb(aehdr0, avhdr2, buf2, m3, boolvar0), apebs[n1] = mkpeb(aechdr(m4), avhdr0, buf0, m1, false), wlar[n0] = wl-entry(m6, used), wlar[n] = wl-entry(m5, erasing), wlar[n1] = wl-entry(m4, used), vols[avhdr.vol][avhdr.leb] = unmapped, apebs[n0] = mkpeb(aechdr(m6), avhdr1, buf1, m2, false), valid-sqns(v, m, apebs) = bag ++ avhdr2.sqn ++ avhdr0.sqn, n < # wlar, avhdr.leb < # vols[avhdr.vol], n0 < # wlar, n1 < # wlar, is-encoded(to-vtbl(vols), buf1), valid-peb(v, m, mkpeb(aehdr0, avhdr2, buf2, m3, boolvar0)), valid-peb(VTBL_VOLID, VTBL_LNUM, mkpeb(aechdr(m6), avhdr1, buf1, m2, false)), valid-peb(mkpeb(aehdr, avhdr, buf, m0, boolvar)), valid-peb(v, m, mkpeb(aechdr(m4), avhdr0, buf0, m1, false)), avhdr.vol ∈ vols, (bag ++ avhdr2.sqn).max ≤ avhdr0.sqn, ∀ v, m, n2. vols[v][m] = ⌜n2⌝ ∧ v ∈ vols ∧ m < # vols[v] ↔ wlar[n2].status = used ∧ n2 < # wlar ∧ apebs[n2].vidhdr.valid? ∧ apebs[n2].vidhdr.vol = v ∧ apebs[n2].vidhdr.leb = m ∧ n2 ≠ n0 ⊦ avhdr0.sqn = (avhdr0.sqn = avhdr2.sqn ⊃ bag ++ avhdr2.sqn;bag ++ avhdr0.sqn).max, n1 = n, VTBL_VOLID ∈ vols, v = avhdr.vol ∧ m = avhdr.leb

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ m = (bag ++ m).max ↔ ¬(bag ≠ ∅ ∧ m < bag.max); (s)% ⊦ bag ≠ ∅ → (bag ++ m).max = (bag.max > m ⊃ bag.max;m); (s)% ⊦ m > n ↔ n < m; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)% ⊦ m < n → (n ≤ m ↔ false); (s)% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ m < n → ¬ n < m; (s)%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic% ⊦ ¬ n < n; (s)