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

⊦ a0 ∈ (s ++ a) \ s1 ↔ a0 = a ∨ a0 ∈ s \ s1, a ∈ s1

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: Difference: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2;%Difference: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2;%In-insert: ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s; from specification set-basic%in-02: ⊦ a ≠ b → (a ∈ s ++ b ↔ a ∈ s); from specification set-basic