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Proof step 2 from proof k-diff-rec1 (previous step)

a ∈ s0 ⊦ s \ s0 = s \ (s0 -- a), a ∈ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
a ∈ s0, a1 ∈ s0 ⊦ a0 ∈ s \ s0 ↔ a0 ∈ s \ (s0 -- a), a ∈ s

Rule argument

Additional information

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