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Proof step 2 from proof valid-eraseq-stronger-prev-append-lref (previous step)

valid-sqns(v, n, apebs, m).max < apebs[m].vidhdr.sqn, m0 < m, m < # apebs, valid-eraseq-stronger(subarray(apebs, 0, m), q), valid-peb(v, n, apebs[m]), valid-peb(v, n, apebs[m0]) ⊦ valid-eraseq-stronger(subarray(apebs, 0, m + 1), q + eq-entry(m0, v × n) '), m0 ∈ q

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
m0 < # apebs, valid-sqns(v, n, apebs, m).max < apebs[m].vidhdr.sqn, m0 < m, m < # apebs, m0 + 1 < # apebs, valid-eraseq-stronger(subarray(apebs, 0, m), q), valid-peb(v, n, apebs[m]), valid-peb(v, n, apebs[m0]) ⊦ valid-eraseq-stronger(subarray(apebs, 0, m + 1), q + eq-entry(m0, v × n) '), m0 ∈ q

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: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic