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Proof step 1 from proof delete-shift (conclusion)

⊦ a ∈ s → (s -- a = s0 ↔ s = s0 ++ a ∧ ¬ a ∈ s0)

System applied rule normalize

And obtained the following new goals

Proof step 2:
a ∈ s ⊦ s -- a = s0 ↔ s = s0 ++ a ∧ ¬ a ∈ s0

Rule argument

Additional information