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Proof step 5 from proof valid-sqns-subseteq-h (previous step)

n < # apebs, n0 ≤ n + 1, n0 ≤ # apebs, ∀ 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 cut formula

And obtained the following new goals

Proof step 6:
n < # apebs, n0 < n, n0 ≤ # 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} ;∅)
Proof step 22:
n < # apebs, n0 ≤ # 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) ⊦ n0 < 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

fma: n0 < n

Additional information

Used formulas: n0 < n