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Proof step 4 from proof lf-valid-recs-max (previous step)

# 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
# apebs = # wlar, apebs[m0] = mkpeb(aehdr, avhdr, buf, m1, boolvar), recs[v × n] = recovery-entry(m0, avhdr.sqn), m0 < m, m0 < # wlar, valid-peb(v, n, mkpeb(aehdr, avhdr, buf, m1, boolvar)), 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

Additional information

Used simplifier rules: lf-valid-recs-sqns-vidhdr: ⊦ v × m ∈ recs ∧ recs[v × m] = recovery-entry(n0, n1) ∧ valid-recs-sqns(apebs, recs) → apebs[n0].vidhdr.sqn = n1;% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).vidhdr = avhdr; (s)