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Proof step 19 from proof valid-ebatbl-update-wlar-unused (previous step)

wlar[m, wl-entry(m1, stat0)][m] = wl-entry(m4, stat1), wlar[m] = wl-entry(m0, stat), apebs[m] = mkpeb(aehdr, avhdr, buf, m3, 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][m2] = ⌜m⌝ ∧ v ∈ vols ∧ m2 < # vols[v] ↔ stat1 = used ∧ m < # apebs ∧ m < # wlar ∧ avhdr.valid? ∧ avhdr.vol = v ∧ avhdr.leb = m2 ∧ m ≠ n, stat = used, stat0 = used, m < # wlar

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
vols[v][m2] = ⌜m⌝ , wlar[m, wl-entry(m1, stat0)][m] = wl-entry(m4, stat1), wlar[m] = wl-entry(m0, stat), apebs[m] = mkpeb(aehdr, avhdr, buf, m3, boolvar), m2 < # vols[v], v ∈ vols, # wlar ≤ m, ∀ 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 ⊦ stat = used, stat0 = used

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n ↔ ¬ n ≤ m; (ss)