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Proof step 9 from proof selsub-diff (previous step)

∀ a. a ∈ s1 → a ∈ s2 ⊦ a ∈ s1 ↔ a ∈ s2, a1 ∈ s1 ↔ a1 ∈ s2 ∧ ¬ a1 ∈ s2 \ s1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
a1 ∈ s1, a0 ∈ s1, ∀ a. a ∈ s1 → a ∈ s2 ⊦ a ∈ s1 ↔ a ∈ s2, a1 ∈ s2, a1 ∈ s2 \ s1

Rule argument

Additional information

Used simplifier rules: Difference: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2;%In-empty: ⊦ ¬ a ∈ ∅; from specification set-basic%Extension: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); from specification set-basic%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic