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Proof step 18 from proof valid-eraseq-stronger-one (previous step)

apebs[m] = mkpeb(aehdr, avhdr, buf, m0, boolvar), lref = v0 × m2, 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(v0, m2, mkpeb(aehdr, avhdr, buf, m0, boolvar)) ∧ avhdr.sqn ≠ valid-sqns(v0, m2, apebs).max

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
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

Rule argument

Additional information

Used simplifier rules: ⊦ none ≠ v × m; (s)