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In lambda calculus, starting from $KII$ we can get by a single $\beta$-reduction to either

$\lambda y.II$

or $KI$

Now, according to Church-Rosser, these should be reducing into a common normal form. But the first reduces to $I$ and the second to $\lambda y.\lambda z.z$.

Am I doing wrong $\beta$-reductions or the theorem holds only on some condition? I do not see a common reduct here.

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    $\begingroup$ (1) You are missing parens in the first term: $(\lambda y. I)I$. (2) How did you get the second term? Remember, $KII$ is syntactic sugar for $(KI)I$. It seems that you did a reduction step as if it was $K(II)$. $\endgroup$ – Anton Trunov Nov 29 '16 at 13:20
  • $\begingroup$ @AntonTrunov and it was indeed K(II) in the book, a momentary lapse of attention. $\endgroup$ – Gergely Dec 13 '16 at 9:19
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Remember that $1)$ application is left associative and $2)$ application has higher precedence than abstraction. This is:

$1)$ $ A B C = (A B) C \neq A (B C)$
$2)$ $ \lambda x. A B = \lambda x. (A B) \neq (\lambda x. A) B$

So

$KII = (KI)I = ((\lambda x y. x) I) I = (\lambda y. I) I = I$

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  • $\begingroup$ sorry my remark was wrong $\endgroup$ – Gergely Nov 30 '17 at 8:38

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