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

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ n0 ≤ n1 ∧ n1 ≤ # apebs → valid-sqns(v, m, apebs, n0) ⊆ valid-sqns(v, m, apebs, n1)

Rule argument

spec: <toplevel> name: valid-sqns-subseteq-h seq: ⊦ n0 ≤ n1 ∧ n1 ≤ # apebs → valid-sqns(v, m, apebs, n0) ⊆ valid-sqns(v, m, apebs, n1) subst: [n1, n0, apebs, m, v] → [# apebs, n, apebs, m, v] flags: <all paths>

Additional information

Used simplifier rules: ⊦ m ≤ m; (s) Used terms: # apebs, n, apebs, m, v