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Proof step 1 from proof sub-insert (conclusion)

⊦ ¬ a ∈ s → (s ⊆ s0 ++ a ↔ s ⊆ s0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
s ⊆ s0 ++ a ⊦ a ∈ s, s ⊆ s0, s0 ++ a ⊆ s0

Rule argument

Additional information

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