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Proof step 1 from proof subset (conclusion)
⊦ list2set(x) ⊆ s ∧ a ∈ x → a ∈ s
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 2
:
a ∈ x, list2set(x) ⊆ s ⊦ a ∈ s
Rule argument
Additional information