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Proof step 21 from proof valid-ebatbl-update-vhdrs-unused (previous step)

wlar[n0] = wl-entry(m3, stat0), wlar[m] = wl-entry(m0, stat), apebs[m, apeb][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 simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
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

Rule argument

Additional information

Used simplifier rules: ⊦ m ≠ n → apebs[n, apeb][m] = apebs[m]; (s)