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Proof step 2 from proof valid-sqns-subseteq (previous step)

n ≤ # apebs ⊦ valid-sqns(v, m, apebs, n) ⊆ valid-sqns(v, m, apebs)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n ≤ # apebs ⊦ valid-sqns(v, m, apebs, n) ⊆ valid-sqns(v, m, apebs, # apebs)

Rule argument

Additional information

Used simplifier rules: valid-sqns: ⊦ valid-sqns(v, m, apebs) = valid-sqns(v, m, apebs, # apebs);