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Proof step 1 from proof subseteq-union-one (conclusion)

n < m, m ≤ # asar ⊦ asar[n] ⊆ union(asar, 0, m)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
n < m, n < # asar, m ≤ # asar ⊦ asar[n] ⊆ union(asar, 0, m)

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic