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Proof step 14 from proof valid-eraseq-stronger-no-maxsqnum (previous step)

avhdr.sqn = valid-sqns(v, m, apebs).max, apebs[m3] = mkpeb(aehdr1, avhdr1, buf1, m4, boolvar1), apebs[n] = mkpeb(aehdr, avhdr, buf, m0, boolvar), n < # apebs, m3 < # apebs, valid-eraseq-stronger(apebs, q0), valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)), ∀ v, m. (valid-peb(v, m, mkpeb(aehdr1, avhdr1, buf1, m4, boolvar1)) ↔ lref = v × m) ∧ (lref = v × m → avhdr1.sqn ≠ valid-sqns(v, m, apebs).max), n ∈ q0 ∨ n = m3, avhdr.sqn ≠ valid-sqns(v, m, apebs).max ∨ apebs[n] ≠ mkpeb(aehdr, avhdr, buf, m0, boolvar) ∨ ¬ n < # apebs ∨ ¬ valid-eraseq-stronger(apebs, q0) ∨ ¬ valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)) ∨ ¬ n ∈ q0 ⊦ m3 ∈ q0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
avhdr1.sqn = valid-sqns(v, m, apebs).max, apebs[m3] = mkpeb(aehdr1, avhdr1, buf1, m4, false), m3 < # apebs, valid-eraseq-stronger(apebs, q0), valid-peb(v, m, mkpeb(aehdr1, avhdr1, buf1, m4, false)), ∀ v0, m1. (v = v0 ∧ m = m1 ↔ lref = v0 × m1) ∧ (lref = v0 × m1 → avhdr1.sqn ≠ valid-sqns(v0, m1, apebs).max) ⊦ m3 ∈ q0

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, true)); (s)%⊦ mkpeb(aehdr, avhdr, buf, m, boolvar) = mkpeb(aehdr0, avhdr0, buf0, m0, boolvar0) ↔ aehdr = aehdr0 ∧ avhdr = avhdr0 ∧ buf = buf0 ∧ m = m0 ∧ (boolvar ↔ boolvar0); (s)%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%⊦ valid-peb(v0, m0, apeb) → (valid-peb(v1, m1, apeb) ↔ v0 = v1 ∧ m0 = m1); (s)