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Proof step 1 from proof del-addeq-04 (conclusion)

⊦ ¬ a ∈ s → (s ++ a) -- a = s

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: del-addeq-05: ⊦ (s ++ a) -- a = s -- a;%del-in: ⊦ ¬ a ∈ s → s -- a = s;%ext: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2);