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Proof step 11 from proof valid-eraseq-stronger-one (previous step)

apebs[m] = mkpeb(aehdr, avhdr, buf, m0, boolvar), m < # apebs ⊦ (∀ v, m1. (valid-peb(v, m1, mkpeb(aehdr, avhdr, buf, m0, boolvar)) ↔ lref = v × m1) ∧ (lref = v × m1 → avhdr.sqn ≠ valid-sqns(v, m1, apebs).max)) ↔ (lref = none → ¬ valid-peb(mkpeb(aehdr, avhdr, buf, m0, boolvar))) ∧ (lref ≠ none → valid-peb(lref.vol, lref.leb, mkpeb(aehdr, avhdr, buf, m0, boolvar)) ∧ avhdr.sqn ≠ valid-sqns(lref.vol, lref.leb, apebs).max)

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

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