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Proof step 24 from proof valid-eraseq-stronger-next-append-lref (previous step)

valid-peb(v0, m1, mkpeb(aehdr0, avhdr0, buf0, m3, boolvar0)) ↔ lref = v0 × m1, subarray(apebs, 0, m + 1)[m2] = mkpeb(aehdr1, avhdr1, buf1, m4, boolvar1), apebs[m2] = mkpeb(aehdr0, avhdr0, buf0, m3, boolvar0), apebs[m] = mkpeb(aehdr, avhdr, buf, m0, boolvar), avhdr.sqn < valid-sqns(v, n, apebs, m).max, m2 < # apebs, m < # apebs, m2 < m, m2 + 1 < # apebs, valid-eraseq-stronger(subarray(apebs, 0, m + 1), q0 + eq-entry(m, v × n) '), valid-eraseq-stronger(subarray(apebs, 0, m), q0), valid-peb(v, n, mkpeb(aehdr, avhdr, buf, m0, boolvar)), lref = v0 × m1 → avhdr0.sqn ≠ valid-sqns(v0, m1, apebs, m).max ⊦ m2 = m, valid-sqns(v, n, apebs, m) = ∅, m2 ∈ q0, (lref = v0 × m1 → avhdr1.sqn ≠ (valid-sqns(v0, m1, apebs, m) ∪ (v = v0 ∧ n = m1 ⊃ {avhdr.sqn} ;∅)).max) ∧ (valid-peb(v0, m1, mkpeb(aehdr1, avhdr1, buf1, m4, boolvar1)) ↔ lref = v0 × m1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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