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Proof step 3 from proof valid-sqns-nonempty-extract-max (previous step)

⊦ valid-sqns(v, m, apebs, # apebs) = ∅, ∃ n. n < # apebs ∧ valid-peb(v, m, apebs[n]) ∧ apebs[n].vidhdr.sqn = valid-sqns(v, m, apebs, # apebs).max

User applied rule apply lemma

And obtained the following new goals

Proof step 4:
⊦ n ≤ # apebs ∧ valid-sqns(v, m, apebs, n) ≠ ∅ → (∃ n0. n0 < n ∧ valid-peb(v, m, apebs[n0]) ∧ apebs[n0].vidhdr.sqn = valid-sqns(v, m, apebs, n).max)
Proof step 5:
# apebs ≤ # apebs ∧ valid-sqns(v, m, apebs, # apebs) ≠ ∅ → (∃ n0. n0 < # apebs ∧ valid-peb(v, m, apebs[n0]) ∧ apebs[n0].vidhdr.sqn = valid-sqns(v, m, apebs, # apebs).max) ⊦ valid-sqns(v, m, apebs, # apebs) = ∅, ∃ n. n < # apebs ∧ valid-peb(v, m, apebs[n]) ∧ apebs[n].vidhdr.sqn = valid-sqns(v, m, apebs, # apebs).max

Rule argument

spec: <toplevel> name: valid-sqns-nonempty-extract-max-h seq: ⊦ n ≤ # apebs ∧ valid-sqns(v, m, apebs, n) ≠ ∅ → (∃ n0. n0 < n ∧ valid-peb(v, m, apebs[n0]) ∧ apebs[n0].vidhdr.sqn = valid-sqns(v, m, apebs, n).max) subst: [n, apebs, v, m] → [# apebs, apebs, v, m]

Additional information

Used terms: # apebs, apebs, v, m