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Proof step 5 from proof subset-is-union (previous step)

¬ a ∈ s1, ∀ s. s ⊆ s1 → (∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅), s ⊆ s1 ++ a ⊦ ∃ s2. s ∪ s2 = s1 ++ a ∧ s ∩ s2 = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information