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

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

User applied rule case distinction

And obtained the following new goals

Proof step 3:
a0 ∈ s1 \ s2 ⊦ a0 ∈ (s1 -- a) \ (s2 -- a), a ∈ s1 \ s2

Rule argument

pos: right 1

Additional information

Used simplifier rules: Difference: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2;%del-rax: ⊦ a ∈ s -- b ↔ ¬ ¬(a ≠ b ∧ a ∈ s);%Delete: ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s;