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Proof step 3 from proof valid-sqns-subseteq-h (previous step)

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

User applied rule structural induction

And obtained the following new goals

Proof step 4:
n0 ≤ # apebs, n0 ≤ n +1, n +1 ≤ # 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 +1)

Rule argument

indvar: n1

Additional information

Used simplifier rules: valid-sqns-base: ⊦ valid-sqns(v, m, apebs, 0) = ∅;%nat freely generated by 0, +1;;% ⊦ 0 ≤ m; (s)% ⊦ m ≤ 0 ↔ m = 0; (s)% ⊦ bag0 ⊆ ∅ ↔ bag0 = ∅; (s)