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Proof step 3 from proof lf-valid-recs-max (previous step)

# apebs = # wlar, recs[v × n] = recovery-entry(m0, apebs[m0].vidhdr.sqn), m0 < m, m0 < # wlar, valid-peb(v, n, apebs[m0]), valid-recs(subarray(apebs, 0, m), subarray(wlar, 0, m), recs), valid-recs-sqns(apebs, recs), valid-wl(subarray(apebs, 0, m), subarray(wlar, 0, m)), v × n ∈ recs, m ≤ # wlar ⊦ apebs[m0].vidhdr.sqn = valid-sqns(v, n, apebs, m).max

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
# apebs = # wlar, recs[v × n] = recovery-entry(m0, avhdr.sqn), apebs[m0] = mkpeb(aehdr, avhdr, buf, m1, boolvar), m0 < # wlar, m0 < m, valid-peb(v, n, apebs[m0]), valid-recs(subarray(apebs, 0, m), subarray(wlar, 0, m), recs), 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: apeb name: elim seq: ⊦ ( ( (aehdr = apeb.echdr ∧ avhdr = apeb.vidhdr) ∧ buf = apeb.data) ∧ m = apeb.fill) ∧ (boolvar ↔ apeb.bad) ↔ apeb = mkpeb(aehdr, avhdr, buf, m, boolvar) subst: [apeb, aehdr, avhdr, buf, m, boolvar] → [apebs[m0], aehdr, avhdr, buf, m1, boolvar]

Additional information

Used simplifier rules: ⊦ ( ((aehdr = apeb.echdr ∧ avhdr = apeb.vidhdr) ∧ buf = apeb.data) ∧ m = apeb.fill) ∧ (boolvar ↔ apeb.bad) ↔ apeb = mkpeb(aehdr, avhdr, buf, m, boolvar); (elim) Used terms: apebs[m0], aehdr, avhdr, buf, m1, boolvar