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Proof step 2 from proof sub-one-01 (previous step)

⊦ s ⊆ ∅ ++ a ↔ s = ∅, s = ∅ ++ a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ a0 ∈ s ↔ a0 = a, (∀ a1. a1 ∈ s → a1 = a) ↔ (∀ a1. ¬ a1 ∈ s)

Rule argument

Additional information

Used simplifier rules: Extension: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2);%in-04: ⊦ a ∈ ∅ ++ b ↔ a = b;%Subset: ⊦ s1 ⊆ s2 ↔ (∀ a. a ∈ s1 → a ∈ s2);%In-empty: ⊦ ¬ a ∈ ∅;