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Proof step 6 from proof valid-eraseq-stronger-no-maxsqnum (previous step)

avhdr.sqn = valid-sqns(v, m, apebs).max, apebs[n] = mkpeb(aehdr, avhdr, buf, m0, boolvar), n < # apebs, valid-eraseq-stronger(apebs, eqent + q0), valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)), n ∈ eqent + q0, ∀ avhdr, v, m, apebs, n, aehdr, buf, m0, boolvar. ¬ ( 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) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
avhdr.sqn = valid-sqns(v, m, apebs).max, apebs[n] = mkpeb(aehdr, avhdr, buf, m0, boolvar), n < # apebs, eqent.pnum < # apebs, valid-eraseq-stronger(apebs, q0), valid-peb(v, m, mkpeb(aehdr, avhdr, buf, m0, boolvar)), n ∈ eqent ' + q0, ∀ v, m. (valid-peb(v, m, apebs[eqent.pnum]) ↔ eqent.lebref = v × m) ∧ (eqent.lebref = v × m → apebs[eqent.pnum].vidhdr.sqn ≠ valid-sqns(v, m, apebs).max), ∀ avhdr0, v0, m1, apebs0, n0, aehdr0, buf0, m2, boolvar0. avhdr0.sqn ≠ valid-sqns(v0, m1, apebs0).max ∨ apebs0[n0] ≠ mkpeb(aehdr0, avhdr0, buf0, m2, boolvar0) ∨ ¬ n0 < # apebs0 ∨ ¬ valid-eraseq-stronger(apebs0, q0) ∨ ¬ valid-peb(v0, m1, mkpeb(aehdr0, avhdr0, buf0, m2, boolvar0)) ∨ ¬ n0 ∈ q0 ⊦ eqent.pnum ∈ q0

Rule argument

Additional information

Used simplifier rules: valid-eraseq-stronger-rec: ⊦ valid-eraseq-stronger(apebs, eqent ' + q) ↔ eqent.pnum < # apebs ∧ ¬ eqent.pnum ∈ q ∧ (∀ v, m. (valid-peb(v, m, apebs[eqent.pnum]) ↔ eqent.lebref = v × m) ∧ ( eqent.lebref = v × m → apebs[eqent.pnum].vidhdr.sqn ≠ valid-sqns(v, m, apebs).max)) ∧ valid-eraseq-stronger(apebs, q);%a: ⊦ (q + q0) + q1 = q + q0 + q1; from specification list[erasequeue-base]% ⊦ eqent + q = eqent ' + q; (s)