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Proof step 13 from proof valid-ebatbl-update-wlar-unused (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 14:
apebs[n0] = mkpeb(aehdr, avhdr, buf, m3, boolvar), wlar[m] = wl-entry(m0, stat), wlar[m, wl-entry(m1, stat0)][n0] = wl-entry(m4, stat1), ∀ 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] = ⌜n0⌝ ∧ v ∈ vols ∧ m2 < # vols[v] ↔ stat1 = used ∧ n0 < # apebs ∧ n0 < # wlar ∧ avhdr.valid? ∧ avhdr.vol = v ∧ avhdr.leb = m2 ∧ n0 ≠ n, stat = used, stat0 = used

Rule argument

spec: wlentry name: elim-02 seq: ⊦ m = wle.ec ∧ stat = wle.status ↔ wle = wl-entry(m, stat) subst: [wle, m, stat] → [wlar[m, wl-entry(m1, stat0)][n0], m4, stat1]

Additional information

Used simplifier rules: ⊦ m = wle.ec ∧ stat = wle.status ↔ wle = wl-entry(m, stat); (elim) Used terms: wlar[m, wl-entry(m1, stat0)][n0], m4, stat1