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Proof step 22 from proof valid-ebatbl-update-vhdrs-unused (previous step)

wlar[m] = wl-entry(m0, stat), wlar[n0] = wl-entry(m3, stat0), apebs[n0] = mkpeb(aehdr, avhdr, buf, m2, boolvar), ∀ v, m, n1. vols[v][m] = ⌜n1⌝ ∧ v ∈ vols ∧ m < # vols[v] ↔ wlar[n1].status = used ∧ n1 < # apebs ∧ n1 < # wlar ∧ apebs[n1].vidhdr.valid? ∧ apebs[n1].vidhdr.vol = v ∧ apebs[n1].vidhdr.leb = m ∧ n1 ≠ n ⊦ vols[v][m1] = ⌜n0⌝ ∧ v ∈ vols ∧ m1 < # vols[v] ↔ stat0 = used ∧ n0 < # apebs ∧ n0 < # wlar ∧ avhdr.valid? ∧ avhdr.vol = v ∧ avhdr.leb = m1 ∧ n0 ≠ n, n0 = m, stat = used

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: v, m1, n0

Additional information

Used simplifier rules: ⊦ wl-entry(m, stat).status = stat; (s)% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).vidhdr = avhdr; (s) Used terms: v, m1, n0