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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m = (bag ++ m).max ↔ ¬(bag ≠ ∅ ∧ m < bag.max); (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%union-com: ⊦ bag ∪ bag2 = bag2 ∪ bag; from specification bag-union[nat-bag]%union-ass: ⊦ bag ∪ bag3 ∪ bag4 = (bag ∪ bag3) ∪ bag4; from specification bag-union[nat-bag]% ⊦ bag ∪ {n} = bag ++ n; (s)%⊦ valid-peb(v0, m0, apeb) → (valid-peb(v1, m1, apeb) ↔ v0 = v1 ∧ m0 = m1); (s)%⊦ n < # apebs → valid-sqns(v, m, apebs, n + 1) = valid-sqns(v, m, apebs, n) ∪ (valid-peb(v, m, apebs[n]) ⊃ {apebs[n].vidhdr.sqn} ;∅); (s)