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Proof step 7 from proof valid-sqns-subseteq-h (previous step)

n0 < # apebs, n < # apebs, n0 < n, n0 + 1 < # apebs, n0 ≤ n + 1, ∀ apebs, n0, v, m. n ≤ # apebs ∧ n0 ≤ n ∧ n0 ≤ # apebs → valid-sqns(v, m, apebs, n0) ⊆ valid-sqns(v, m, apebs, n) ⊦ valid-sqns(v, m, apebs, n0) ⊆ valid-sqns(v, m, apebs, n) ∪ (valid-peb(v, m, apebs[n]) ⊃ {apebs[n].vidhdr.sqn} ;∅)

User applied rule all left

And obtained the following new goals

Proof step 8:
n0 + 1 < # apebs, n0 < n, n < # apebs, n0 < # apebs, n0 ≤ n + 1, n ≤ # apebs ∧ n0 ≤ n ∧ n0 ≤ # apebs → valid-sqns(v, m, apebs, n0) ⊆ valid-sqns(v, m, apebs, n) ⊦ valid-sqns(v, m, apebs, n0) ⊆ valid-sqns(v, m, apebs, n) ∪ (valid-peb(v, m, apebs[n]) ⊃ {apebs[n].vidhdr.sqn} ;∅)

Rule argument

pos: left 6 terms: apebs, n0, v, m abort: F discard: T computed: F

Additional information

Used terms: apebs, n0, v, m