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Proof step 2 from proof valid-sqns-extract2-sqn-h (previous step)

n0 < m0, n1 < m0, valid-peb(v, m, apebs[n0]), valid-peb(v, m, apebs[n1]), m0 ≤ # apebs ⊦ n0 = n1, ∃ bag. valid-sqns(v, m, apebs, m0) = bag ++ apebs[n0].vidhdr.sqn ++ apebs[n1].vidhdr.sqn

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n1 < m0, n0 < m0, n1 < # apebs, n0 < # apebs, valid-peb(v, m, apebs[n1]), valid-peb(v, m, apebs[n0]), m0 ≤ # apebs ⊦ n0 = n1, ∃ bag. valid-sqns(v, m, apebs, m0) = bag ++ apebs[n0].vidhdr.sqn ++ apebs[n1].vidhdr.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