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| |} | | |} |
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− | <pre> | + | Observe that <math>y(xz)\!</math> matches <math>(xy)(xz)\!</math> on the right, and that we can express <math>y\!</math> as <math>x(y\operatorname{K}),</math> consequently: |
− | Observe that y(xz) matches (xy)(xz) on the right,
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− | and that we can express y as x(yK), consequently: | |
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− | y(xz) = (x(yK))(xz)
| + | {| align="center" cellpadding="8" width="90%" |
| + | | |
| + | <math>\begin{matrix} |
| + | y(xz) |
| + | & = & |
| + | (x(y\operatorname{K}))(xz) |
| + | \\[8pt] |
| + | & = & |
| + | x((y\operatorname{K})(z\operatorname{S})) |
| + | \end{matrix}</math> |
| + | |} |
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− | = x((yK)(zS))
| + | thus completing the abstraction (or disentanglement) of x from the expression. |
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− | thus completing the abstraction (or disentanglement)
| + | Working on the remainder of the expression, the next item of business is to abstract <math>y.\!</math> |
− | of x from the expression.
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− | | |
− | Working on the remainder of the expression, | |
− | the next item of business is to abstract y. | |
| | | |
| + | <pre> |
| Notice that: | | Notice that: |
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