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Proof step 2 from proof union-subset-02 (previous step)

s ⊆ s0, s1 ⊆ s2 ⊦ s ∪ s1 ⊆ (s0 ∪ s2) ++ a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ a. a ∈ s1 → a ∈ s2, ∀ a. a ∈ s → a ∈ s0, a0 ∈ s ∨ a0 ∈ s1 ⊦ a0 = a, a0 ∈ s0, a0 ∈ s2

Rule argument

Additional information

Used simplifier rules: unrax: ⊦ a ∈ s ∪ s0 ↔ a ∈ s ∨ a ∈ s0;%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;% ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s; (ws)% ⊦ s1 ⊆ s2 ↔ (∀ a. a ∈ s1 → a ∈ s2); (s)