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Proof step 1 from proof del-addeq-06 (conclusion)

⊦ (s ++ a ++ b) -- a = (s ++ b) -- a

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: Delete: ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s;% ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s; (s)% ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); (s)