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"union" or "disjunction" in pure untyped lambda calculus

The usual way to encode coproducts is as: $$ι_1\ x = λk_1\ k_2.\ k_1\ x \\ ι_2\ y = λk_1\ k_2.\ k_2\ y$$ Matching is then, as you said, $[f,g]= λs.\ s\ f\ g$. The obvious problem with idempotence is ...
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