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Proof step 19 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)) ↔ v0 × m2 = v × m1) ∧ (v0 × m2 = v × m1 → avhdr.sqn ≠ valid-sqns(v, m1, apebs).max) ⊦ valid-peb(v0, m2, mkpeb(aehdr, avhdr, buf, m0, boolvar)) ∧ avhdr.sqn ≠ valid-sqns(v0, m2, 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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