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Proof step 8 from proof sub-insert (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
∀ a0. a0 ∈ s → a0 ∈ s0 ⊦ a ∈ s0, a ∈ s, s ⊆ s0, s0 ++ a ⊆ s0

Rule argument

Additional information

Used simplifier rules: rax: ⊦ a ∈ s ++ b ↔ ¬ ¬(a = b ∨ a ∈ s);