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Proof step 10 from proof lf-valid-recs-max (previous step)

# apebs = # wlar, wlar[m0].status = used, recs[v × n] = recovery-entry(m0, avhdr.sqn), apebs[m0] = mkpeb(aehdr, avhdr, buf, m1, boolvar), m0 < # wlar, m0 < m, valid-peb(v, n, mkpeb(aehdr, avhdr, buf, m1, boolvar)), valid-recs-sqns(apebs, recs), valid-wl(subarray(apebs, 0, m), subarray(wlar, 0, m)), v × n ∈ recs, m ≤ # wlar ⊦ avhdr.sqn = valid-sqns(v, n, apebs, m).max

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 11:
# apebs = # wlar, apebs[m0] = mkpeb(aehdr, avhdr, buf, m1, boolvar), recs[v × n] = recovery-entry(m0, avhdr.sqn), wlar[m0] = wl-entry(m2, stat), stat = used, m0 < m, m0 < # wlar, valid-peb(v, n, mkpeb(aehdr, avhdr, buf, m1, boolvar)), valid-recs-sqns(apebs, recs), valid-wl(subarray(apebs, 0, m), subarray(wlar, 0, m)), v × n ∈ recs, m ≤ # wlar ⊦ avhdr.sqn = valid-sqns(v, n, apebs, m).max

Rule argument

spec: wlentry name: elim-02 seq: ⊦ m = wle.ec ∧ stat = wle.status ↔ wle = wl-entry(m, stat) subst: [wle, m, stat] → [wlar[m0], m2, stat]

Additional information

Used simplifier rules: ⊦ m = wle.ec ∧ stat = wle.status ↔ wle = wl-entry(m, stat); (elim) Used terms: wlar[m0], m2, stat