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Proof step 8 from proof valid-global-sqn-add-valid (previous step)

avhdr.sqn < ( (valid-sqns(v, m0, apebs) \ (valid-peb(v, m0, apebs[m]) ⊃ {apebs[m].vidhdr.sqn} ;∅)) ∪ (valid-peb(v, m0, mkpeb(aehdr, avhdr, buf, m1, boolvar)) ⊃ {mkpeb(aehdr, avhdr, buf, m1, boolvar).vidhdr.sqn} ;∅) ).max, avhdr.sqn < n, m < # apebs, valid-peb(mkpeb(aehdr, avhdr, buf, m1, boolvar)), n ≤ ( (valid-sqns(v, m0, apebs) \ (valid-peb(v, m0, apebs[m]) ⊃ {apebs[m].vidhdr.sqn} ;∅)) ∪ (valid-peb(v, m0, mkpeb(aehdr, avhdr, buf, m1, boolvar)) ⊃ {mkpeb(aehdr, avhdr, buf, m1, boolvar).vidhdr.sqn} ;∅) ).max, valid-sqns(v, m0, apebs) ≠ ∅ → valid-sqns(v, m0, apebs).max < n ⊦ (valid-sqns(v, m0, apebs) \ (valid-peb(v, m0, apebs[m]) ⊃ {apebs[m].vidhdr.sqn} ;∅)) ∪ (valid-peb(v, m0, mkpeb(aehdr, avhdr, buf, m1, boolvar)) ⊃ {mkpeb(aehdr, avhdr, buf, m1, boolvar).vidhdr.sqn} ;∅) = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
avhdr.sqn < ( (valid-sqns(v, m0, apebs) \ (valid-peb(v, m0, apebs[m]) ⊃ {apebs[m].vidhdr.sqn} ;∅)) ∪ (valid-peb(v, m0, mkpeb(aehdr, avhdr, buf, m1, boolvar)) ⊃ {avhdr.sqn} ;∅) ).max, avhdr.sqn < n, m < # apebs, valid-peb(mkpeb(aehdr, avhdr, buf, m1, boolvar)), n ≤ ( (valid-sqns(v, m0, apebs) \ (valid-peb(v, m0, apebs[m]) ⊃ {apebs[m].vidhdr.sqn} ;∅)) ∪ (valid-peb(v, m0, mkpeb(aehdr, avhdr, buf, m1, boolvar)) ⊃ {avhdr.sqn} ;∅) ).max, valid-sqns(v, m0, apebs) ≠ ∅ → valid-sqns(v, m0, apebs).max < n ⊦ (valid-sqns(v, m0, apebs) \ (valid-peb(v, m0, apebs[m]) ⊃ {apebs[m].vidhdr.sqn} ;∅)) ∪ (valid-peb(v, m0, mkpeb(aehdr, avhdr, buf, m1, boolvar)) ⊃ {avhdr.sqn} ;∅) = ∅

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ mkpeb(aehdr, avhdr, buf, m, boolvar).vidhdr = avhdr; (s)