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Proof step 4 from proof sub-insert (previous step)

⊦ s ⊆ s0 ++ a ↔ s ⊆ s0, a ∈ s, a ∈ s0

User applied rule case distinction

And obtained the following new goals

Proof step 5:
s ⊆ s0 ++ a ⊦ a ∈ s0, a ∈ s, s ⊆ s0

Rule argument

pos: right 1

Additional information

Used simplifier rules: sub-ntrans: ⊦ s ⊆ s0 ∧ ¬ s ⊆ s1 → ¬ s0 ⊆ s1;%sub-add-01: ⊦ s ⊆ s ++ a;