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Proof step 6 from proof valid-eraseq-stronger-imp-strong (previous step)

eqent = eq-entry(m0, lref), m0 < # apebs, valid-eraseq-stronger(apebs, q0), ∀ apebs, vtbl. valid-eraseq-stronger(apebs, q0) → valid-eraseq-strong(apebs, vtbl, q0), ∀ v, m. (valid-peb(v, m, apebs[m0]) ↔ lref = v × m) ∧ (lref = v × m → apebs[m0].vidhdr.sqn ≠ valid-sqns(v, m, apebs).max) ⊦ m0 ∈ q0, (valid-peb(v, m, apebs[m0]) ↔ lref = v × m) ∧ (lref = v × m ∧ v ∈ vtbl ∧ m < vtbl[v] → apebs[m0].vidhdr.sqn ≠ valid-sqns(v, m, apebs).max) ∧ valid-eraseq-strong(apebs, vtbl, q0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
m0 < # apebs, valid-eraseq-stronger(apebs, q0), ∀ v, m. (valid-peb(v, m, apebs[m0]) ↔ lref = v × m) ∧ (lref = v × m → apebs[m0].vidhdr.sqn ≠ valid-sqns(v, m, apebs).max), ∀ apebs, vtbl. valid-eraseq-stronger(apebs, q0) → valid-eraseq-strong(apebs, vtbl, q0) ⊦ m0 ∈ q0, (valid-peb(v, m, apebs[m0]) ↔ lref = v × m) ∧ (lref = v × m ∧ v ∈ vtbl ∧ m < vtbl[v] → apebs[m0].vidhdr.sqn ≠ valid-sqns(v, m, apebs).max) ∧ valid-eraseq-strong(apebs, vtbl, q0)

Proof step 6 from proof valid-eraseq-stronger-imp-strong (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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