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Proof step 1 from proof selsub-diff (conclusion)

s1 ⊆ s2 ⊦ s1 = s2, ∃ s. s ⊆ s2 ∧ s1 = s2 \ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
∀ a. a ∈ s1 → a ∈ s2 ⊦ a ∈ s1 ↔ a ∈ s2, ∃ s. (∀ a. a ∈ s → a ∈ s2) ∧ (∀ a. a ∈ s1 ↔ a ∈ s2 \ s)

Rule argument

Additional information

Used simplifier rules: ⊦ s1 ⊆ s2 ↔ (∀ a. a ∈ s1 → a ∈ s2); (s)% ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); (s)