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Proof step 4 from proof valid-recs-in-nonempty-sqns (previous step)

# wlar = # apebs, v × n ∈ recs ∧ recs[v × n].pnum = recs[v × n].pnum ↔ wlar[recs[v × n].pnum].status = used ∧ recs[v × n].pnum < m ∧ apebs[recs[v × n].pnum].vidhdr.vol = (v × n).vol ∧ apebs[recs[v × n].pnum].vidhdr.leb = (v × n).leb, valid-sqns(v, n, apebs, m) = ∅, m < # apebs, valid-wl(subarray(apebs, 0, m), subarray(wlar, 0, m)), v × n ∈ recs ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
# wlar = # apebs, apebs[recs[v × n].pnum].vidhdr.leb = n, wlar[recs[v × n].pnum].status = used, apebs[recs[v × n].pnum].vidhdr.vol = v, valid-sqns(v, n, apebs, m) = ∅, recs[v × n].pnum + 1 < # apebs, m < # apebs, recs[v × n].pnum < m, recs[v × n].pnum < # apebs, valid-wl(subarray(apebs, 0, m), subarray(wlar, 0, m)), v × n ∈ recs ⊦

Rule argument

Additional information

Used simplifier rules: lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ m +1 = m + 1; (s)% ⊦ m < n → n ≠ 0; (s)%transitivity: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification oelem[nat-set]%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ (v × m).leb = m; (s)% ⊦ (v × m).vol = v; (s)