Definitions lambda jlc Sections Support(jlc) Doc

No mentions to report in lambda_jlc
applicative_YDef Y(f) == (x.f(y.x(x,y)))(x.f(y.x(x,y)))

Syntax:Y has structure: applicative_Y

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lambdaapply!abstraction

Definitions lambda jlc Sections Support(jlc) Doc