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Proof step 9 from proof prepost-get-pebs-for-wl (previous step)

# apebs = # wlar, wlar0 = wlar, bflipset-cons(bflipset, wlar), unique-sqns(apebs), valid-ebatbl(apebs, wlar, vols, vtblpnum1), valid-eraseq(apebs, q), valid-eraseq(wlar, q), valid-eraseq(q), valid-global-sqn(apebs, sqnum), valid-vtbl(apebs, vols, wlar, vtblpnum1), valid-wl(apebs, wlar) ⊦ [ubi_wl_get_free_min#(wlar; n; boolvar)] ( (boolvar → n = free-ecs(wlar0).min) ∧ (¬ boolvar ↔ free-ecs(wlar0) = ∅) → ⟪boolvar0 := false; bflipset0 := bflipset; 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[vtblpnum].ec + WL_THRESHOLD ≤ wlar[vtblpnum0].ec 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)))), VTBL_VOLID ∈ vols, vtblpnum1 ∈ q

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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