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Proof step 11 from proof sub-insert (previous step)

∀ a0. a0 ∈ s → a0 ∈ s0 ⊦ a ∈ s0, a ∈ s, ∀ a. a ∈ s → a ∈ s0, s0 ++ a ⊆ s0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
a0 ∈ s, ∀ a0. a0 ∈ s → a0 ∈ s0 ⊦ s0 ++ a ⊆ s0, a0 ∈ s0, a ∈ s, a ∈ s0

Rule argument

Additional information

Used simplifier rules: ne-01: ⊦ a ∈ s → s ≠ ∅;%ne-02: ⊦ a ∈ s → (s = ∅ ↔ false);