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

n ≤ # apebs ⊦ valid-sqns(v, m, subarray(apebs, 0, n), n) = valid-sqns(v, m, apebs, n)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
n ≤ # apebs, n0 ≤ n ⊦ valid-sqns(v, m, subarray(apebs, 0, n), n0) = valid-sqns(v, m, apebs, n0)

Rule argument

spec: <toplevel> name: valid-sqns-subarray-h seq: n ≤ # apebs, n0 ≤ n ⊦ valid-sqns(v, m, subarray(apebs, 0, n), n0) = valid-sqns(v, m, apebs, n0) subst: [n0, n, apebs, m, v] → [n, n, apebs, m, v] flags: <all paths>

Additional information

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