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Proof step 2 from proof union-subset (previous step)

⊦ union(asar, n, m) ⊆ as ↔ (∀ n0. n ≤ n0 ∧ n0 < n + m → asar[n0] ⊆ as)

User applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ asar, n, as. union(asar, n, m0) ⊆ as ↔ (∀ n0. n ≤ n0 ∧ n0 < n + m0 → asar[n0] ⊆ as) ⊦ union(asar, n, m0 +1) ⊆ as ↔ (∀ n0. n ≤ n0 ∧ n0 < n + m0 +1 → asar[n0] ⊆ as)

Rule argument

indvar: m

Additional information

Used simplifier rules: union-base: ⊦ union(asar, n, 0) = ∅;%nat freely generated by 0, +1;;% ⊦ ∅ ⊆ as; (s)% ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m ≤ n → (n < m ↔ false); (s)