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

n0 ≤ n1, n1 ≤ # apebs ⊦ valid-sqns(v, m, apebs, n0) ⊆ valid-sqns(v, m, apebs, n1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n1 ≤ # apebs, n0 ≤ n1, n0 ≤ # apebs ⊦ valid-sqns(v, m, apebs, n0) ⊆ valid-sqns(v, m, apebs, n1)

Rule argument

Additional information

Used simplifier rules: let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat