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Proof step 1 from proof del-addeq-03 (conclusion)

⊦ a ∈ s ∧ a ≠ b → (s -- b) ++ a = s -- b

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: del-in-01: ⊦ a ≠ b → (¬ a ∈ s -- b ↔ ¬ a ∈ s);%eq-03: ⊦ s ++ a = s ↔ a ∈ s; from specification set-basic