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Proof step 5 from proof selsub-diff (previous step)

a0 ∈ s1 → a0 ∈ s2, ∀ a. a ∈ s1 → a ∈ s2 ⊦ a ∈ s1 ↔ a ∈ s2, (a0 ∈ s2 \ s1 → a0 ∈ s2) ∧ (a1 ∈ s1 ↔ a1 ∈ s2 \ s2 \ s1)

User applied rule case distinction

And obtained the following new goals

Proof step 6:
a0 ∈ s1 → a0 ∈ s2, ∀ a. a ∈ s1 → a ∈ s2 ⊦ a1 ∈ s1 ↔ a1 ∈ s2 \ s2 \ s1, a ∈ s1 ↔ a ∈ s2

Rule argument

pos: right 2

Additional information

Used simplifier rules: Difference: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2;