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Proof step 5 from proof prepost-get-pebs-for-wl (previous step)

# apebs = # wlar, bflipset0 = bflipset, vtblpnum = m, vtblpnum0 = n, bflipset-cons(bflipset, wlar), unique-sqns(apebs), valid-ebatbl(apebs, wlar, vols, vtblpnum1), valid-eraseq(q), valid-eraseq(wlar, q), valid-eraseq(apebs, q), valid-global-sqn(apebs, sqnum), valid-vtbl(apebs, vols, wlar, vtblpnum1), valid-wl(apebs, wlar) ⊦ ⟪let boolvar = ?, n = ? in {ubi_wl_get_free_min#(wlar; n; boolvar); boolvar0 := false; if boolvar then {ubi_wl_find_free_max_below#(n + 2 * WL_THRESHOLD, wlar; vtblpnum0); if bflipset ≠ ∅ then choose n with n ∈ bflipset in {vtblpnum := n; bflipset := bflipset -- n; boolvar0 := true} else {ubi_wl_get_used_min#(wlar; vtblpnum; boolvar); if boolvar ∧ wlar[vtblpnum0].ec ≥ wlar[vtblpnum].ec + WL_THRESHOLD then boolvar0 := true}}}⟫ ( bflipset-cons(bflipset, wlar) ∧ ( boolvar0 → vtblpnum0 < # wlar ∧ wlar[vtblpnum0].status = free ∧ vtblpnum < # wlar ∧ wlar[vtblpnum].status = used ∧ (bflipset0 = ∅ → wlar[vtblpnum].ec + WL_THRESHOLD ≤ wlar[vtblpnum0].ec))), vtblpnum1 ∈ q, VTBL_VOLID ∈ vols

Heuristic symbolic execution applied rule let right

And obtained the following new goals

Rule argument

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