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Proof step 3 from proof valid-sqns-extract-sqn-h (previous step)

n < m0, n < # apebs, valid-peb(v, m, apebs[n]), m0 ≤ # apebs ⊦ ∃ bag. valid-sqns(v, m, apebs, m0) = bag ++ apebs[n].vidhdr.sqn

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 4:
n < # apebs, n < m1 +1, valid-peb(v, m, apebs[n]), m1 +1 ≤ # apebs, ∀ n, apebs, v, m. n < m1 ∧ n < # apebs ∧ valid-peb(v, m, apebs[n]) ∧ m1 ≤ # apebs → (∃ bag. valid-sqns(v, m, apebs, m1) = bag ++ apebs[n].vidhdr.sqn) ⊦ ∃ bag. valid-sqns(v, m, apebs, m1 +1) = bag ++ apebs[n].vidhdr.sqn

Rule argument

term: m0

Additional information

Used simplifier rules: nat freely generated by 0, +1;;% ⊦ 0 ≤ m; (s)% ⊦ ¬ m < 0; (s)