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Proof step 5 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)), (n < m1 + 1 ⊃ m1 + 1;n) ≤ (valid-sqns(v0, m5, apebs, m) ∪ (valid-peb(v0, m5, mkpeb(aehdr, avidhdr(v, m0, m1, m2, m4), buf, m3, false)) ⊃ {m1} ;∅)).max, ∀ v0, m5. valid-sqns(v0, m5, apebs, m) ≠ ∅ → valid-sqns(v0, m5, apebs, m).max < n ⊦ valid-sqns(v0, m5, apebs, m) ∪ (valid-peb(v0, m5, mkpeb(aehdr, avidhdr(v, m0, m1, m2, m4), buf, m3, false)) ⊃ {m1} ;∅) = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
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 < n ⊃ m1 + 1;n) ≤ (valid-sqns(v0, m5, apebs, m) ∪ (valid-peb(v0, m5, mkpeb(aehdr, avidhdr(v, m0, m1, m2, m4), buf, m3, false)) ⊃ {m1} ;∅)).max, ∀ v0, m5. valid-sqns(v0, m5, apebs, m) ≠ ∅ → valid-sqns(v0, m5, apebs, m).max < n ⊦ valid-sqns(v0, m5, apebs, m) ∪ (valid-peb(v0, m5, mkpeb(aehdr, avidhdr(v, m0, m1, m2, m4), buf, m3, false)) ⊃ {m1} ;∅) = ∅

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)