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Proof step 2 from proof diff-del-01 (previous step)

⊦ a ∈ s, s \ s0 ⊆ (s \ s0) -- a

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: Delete: ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s;%Difference: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2;% ⊦ s1 ⊆ s2 ↔ (∀ a. a ∈ s1 → a ∈ s2); (s)