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Proof step 2 from proof subseteq-union-diff-one (previous step)

n0 < # asar, n < # asar, disj(asar, 0, # asar) ⊦ asar[n] ⊆ (union(asar, 0, # asar) \ asar[n0]) ∪ as ↔ n ≠ n0 ∨ asar[n0] ⊆ as

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n0 < # asar, n < # asar, disj(asar, 0, # asar) ⊦ asar[n] ⊆ (union(asar, 0, # asar) \ asar[n0]) ∪ as ↔ n ≠ n0 ∨ asar[n0] ⊆ as

Rule argument

Additional information

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