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Proof step 1 from proof delete-shift (conclusion)

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

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: del-addeq-02: ⊦ a ∈ s → (s -- a) ++ a = s;%del-rax: ⊦ a ∈ s -- b ↔ ¬ ¬(a ≠ b ∧ a ∈ s);%del-addeq-04: ⊦ ¬ a ∈ s → (s ++ a) -- a = s;