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Proof step 7 from proof selsub-diff (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
a1 ∈ s1, a2 ∈ s1, ∀ a. a ∈ s1 → a ∈ s2, a0 ∈ s1 → a0 ∈ s2 ⊦ a ∈ s1 ↔ a ∈ s2, a1 ∈ s2 \ s2 \ s1

Rule argument

Additional information

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