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Proof step 5 from proof sub-insert (previous step)

s ⊆ s0 ++ a ⊦ a ∈ s0, a ∈ s, s ⊆ s0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
s ⊆ s0 ++ a ⊦ a ∈ s, a ∈ s0, s ⊆ s0, s0 ++ a ⊆ s0

Rule argument

Additional information

Used simplifier rules: sub-ntrans: ⊦ s ⊆ s0 ∧ ¬ s ⊆ s1 → ¬ s0 ⊆ s1;