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Proof step 5 from proof ubi-is-max-unique (previous step)

is-max(apebs, vols0, wlar0, q0, bflipset0, n0, n1), is-max(apebs, vols1, wlar1, q1, bflipset1, m0, m1), is-max(apebs, vols0, wlar0, q0, bflipset0, n0, n1) ∧ ubi-cons(apebs, vols1, wlar1, q1, bflipset1, m0, m1) → vols1 ⊆ vols0 ⊦ vols0 = vols1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
is-max(apebs, vols1, wlar1, q1, bflipset1, m0, m1), is-max(apebs, vols0, wlar0, q0, bflipset0, n0, n1), # apebs = # wlar1 ∧ unique-sqns(apebs) ∧ valid-global-sqn(apebs, m1) ∧ valid-wl(apebs, wlar1) ∧ valid-eraseq(q1) ∧ valid-eraseq(wlar1, q1) ∧ valid-eraseq(apebs, q1) ∧ ¬ VTBL_VOLID ∈ vols1 ∧ bflipset-cons(bflipset1, wlar1) ∧ valid-ebatbl(apebs, wlar1, vols1, m0) ∧ valid-vtbl(apebs, vols1, wlar1, m0) → vols1 ⊆ vols0 ⊦ vols0 = vols1

Rule argument

Additional information

Used simplifier rules: ⊦ ubi-cons(apebs, vols, wlar, q, bflipset, n0, n1) ↔ # apebs = # wlar ∧ unique-sqns(apebs) ∧ valid-global-sqn(apebs, n1) ∧ valid-wl(apebs, wlar) ∧ valid-eraseq(q) ∧ valid-eraseq(wlar, q) ∧ valid-eraseq(apebs, q) ∧ ¬ VTBL_VOLID ∈ vols ∧ bflipset-cons(bflipset, wlar) ∧ valid-ebatbl(apebs, wlar, vols, n0) ∧ valid-vtbl(apebs, vols, wlar, n0); (s)