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Proof step 2 from proof vols-is-max-is-max-vols (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
# apebs = # wlar1, bflipset-cons(bflipset1, wlar1), is-max(apebs, vols0, wlar0, q0, bflipset0, n0, n1), unique-sqns(apebs), valid-ebatbl(apebs, wlar1, vols1, m0), valid-eraseq(q1), valid-eraseq(wlar1, q1), valid-eraseq(apebs, q1), valid-global-sqn(apebs, m1), valid-vtbl(apebs, vols1, wlar1, m0), valid-wl(apebs, wlar1) ⊦ VTBL_VOLID ∈ vols1, (v ∈ vols1 ↔ v ∈ vols0) ∧ (v ∈ vols1 → # vols1[v] = # vols0[v] ∧ (m < # vols1[v] ∧ vols1[v][m] ≠ unmapped → vols1[v][m] = vols0[v][m]))

Rule argument

Additional information

Used simplifier rules: vols-le: ⊦ vols0 ⊆ vols1 ↔ (∀ v. (v ∈ vols0 ↔ v ∈ vols1) ∧ ( v ∈ vols0 → # vols0[v] = # vols1[v] ∧ (∀ m. m < # vols0[v] ∧ vols0[v][m] ≠ unmapped → vols0[v][m] = vols1[v][m])));%⊦ 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)