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Proof step 3 from proof union-subset (previous step)

∀ 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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ asar, n, as. union(asar, n, m0) ⊆ as ↔ (∀ n0. n ≤ n0 ∧ n0 < n + m0 → asar[n0] ⊆ as) ⊦ asar[n + m0] ∪ union(asar, n, m0) ⊆ as ↔ (∀ n0. n ≤ n0 ∧ ¬ n + m0 < n0 → asar[n0] ⊆ as)

Rule argument

Additional information

Used simplifier rules: union-rec: ⊦ union(asar, n, m + 1) = asar[n + m] ∪ union(asar, n, m);% ⊦ m < n + 1 ↔ ¬ n < m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m +1 = m + 1; (s)