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Proof step 20 from proof valid-global-sqn-next-max (previous step)

apebs[m] = mkpeb(aehdr, avidhdr(v, m0, m1, m2, m4), buf, m3, false), m < # apebs, valid-peb(mkpeb(aehdr, avidhdr(v, m0, m1, m2, m4), buf, m3, false)), m1 + 1 ≤ valid-sqns(v0, m5, apebs, m).max, ∀ v0, m5. valid-sqns(v0, m5, apebs, m) ≠ ∅ → valid-sqns(v0, m5, apebs, m).max < n ⊦ valid-sqns(v0, m5, apebs, m) = ∅, m1 < n, v0 = v ∧ m5 = m0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
apebs[m] = mkpeb(aehdr, avidhdr(v, m0, m1, m2, m4), buf, m3, false), n < valid-sqns(v0, m5, apebs, m).max, m1 < valid-sqns(v0, m5, apebs, m).max, m < # apebs, valid-peb(mkpeb(aehdr, avidhdr(v, m0, m1, m2, m4), buf, m3, false)), n ≤ m1, ∀ v0, m5. valid-sqns(v0, m5, apebs, m) ≠ ∅ → valid-sqns(v0, m5, apebs, m).max < n ⊦ valid-sqns(v0, m5, apebs, m) = ∅, v0 = v ∧ m5 = m0

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)