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Proof step 13 from proof lf-mapped-valid-peb (previous step)

avhdr.sqn = valid-sqns(v0, m1, apebs).max, vols[v][m] = ⌜n⌝ , wlar[n] = wl-entry(m0, used), apebs[n] = mkpeb(aechdr(m0), avhdr, buf, m2, false), n < # wlar, m < # vols[v], valid-peb(v0, m1, mkpeb(aechdr(m0), avhdr, buf, m2, false)), v ∈ vols, ∀ v, m, n. vols[v][m] = ⌜n⌝ ∧ v ∈ vols ∧ m < # vols[v] ↔ wlar[n].status = used ∧ n < # apebs ∧ n < # wlar ∧ apebs[n].vidhdr.valid? ∧ apebs[n].vidhdr.vol = v ∧ apebs[n].vidhdr.leb = m ∧ n ≠ n0 ⊦ n = n0, v0 = v ∧ m1 = m

User applied rule all left

And closed the goal!

Rule argument

pos: left 9 terms: v, m, n abort: F discard: T computed: T

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic% ⊦ wl-entry(m, stat).status = stat; (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).vidhdr = avhdr; (s)%⊦ valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)) → avhdr.vol = v; (s)%⊦ valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)) → avhdr.leb = m; (s)% ⊦ m < n → n ≠ 0; (s) Used terms: v, m, n