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

∀ a. a ∈ s1 → a ∈ s2 ⊦ a ∈ s1 ↔ a ∈ s2, ∃ s. (∀ a. a ∈ s → a ∈ s2) ∧ (∀ a. a ∈ s1 ↔ a ∈ s2 \ s)

User applied rule exists right

And obtained the following new goals

Proof step 3:
∀ a. a ∈ s1 → a ∈ s2 ⊦ a ∈ s1 ↔ a ∈ s2, (∀ a. a ∈ s2 \ s1 → a ∈ s2) ∧ (∀ a. a ∈ s1 ↔ a ∈ s2 \ s2 \ s1)

Rule argument

pos: right 2 terms: s2 \ s1 abort: F discard: T computed: F

Additional information

Used terms: s2 \ s1