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Proof step 4 from proof diff-del (previous step)

⊦ a0 ∈ s1 \ s2 ↔ a0 ∈ (s1 -- a) \ (s2 -- a), a ∈ s1 \ s2, a ∈ s1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ a0 ∈ s1 \ s2 ↔ a0 ∈ s1 \ (s2 -- a), a ∈ s1, a ∈ s1 \ s2

Rule argument

Additional information

Used simplifier rules: del-in: ⊦ ¬ a ∈ s → s -- a = s;