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(λg.(λf.((λx.(f (x x)))(λx.(f (x x))))) g)
λg.((λx.(g (x x)))(λx.(g (x x))))
λg.((λx.(g x x))(λx.(g x x)))
λg.((g λx.(g x x) λx.(g x x)))
//Now this falls in recursion which makes it can't solve
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Lambda Calculus - Reductions #4
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