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Proof step 37 from proof valid-eraseq-stronger-prev-append-lref (previous step)

avhdr0.sqn = avhdr.sqn, apebs[m3] = mkpeb(aehdr0, avhdr0, buf0, m4, false), apebs[m] = mkpeb(aehdr, avhdr, buf, m1, boolvar), apebs[m0] = mkpeb(aehdr1, avhdr1, buf1, m5, boolvar0), valid-sqns(v0, m2, apebs, m) = bag ++ avhdr1.sqn ++ avhdr.sqn, (bag ++ avhdr1.sqn ++ avhdr.sqn).max + 1 < (bag ++ avhdr1.sqn).max, avhdr.sqn < (bag ++ avhdr1.sqn).max, (bag ++ avhdr1.sqn ++ avhdr.sqn).max < avhdr.sqn, m3 + 1 < # apebs, m3 < m, m0 < # apebs, m0 < m, m < # apebs, m0 + 1 < # apebs, m3 < # apebs, (bag ++ avhdr1.sqn ++ avhdr.sqn).max < (bag ++ avhdr1.sqn).max, valid-eraseq-stronger(subarray(apebs, 0, m + 1), q0 + eq-entry(m0, v0 × m2) '), valid-eraseq-stronger(subarray(apebs, 0, m), q0), valid-peb(v0, m2, mkpeb(aehdr, avhdr, buf, m1, boolvar)), valid-peb(v0, m2, apebs[m0]), valid-peb(v0, m2, mkpeb(aehdr0, avhdr0, buf0, m4, false)) ⊦ m0 = m3, m3 ∈ q0, m0 ∈ q0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)% ⊦ m < n → n ≠ 0; (s)%transitivity: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification oelem[nat-set]%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic