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| The negation <math>\lnot x</math> is written <math>(x).\!</math> | | The negation <math>\lnot x</math> is written <math>(x).\!</math> |
| + | |
| + | This corresponds to the logical graph: |
| + | |
| + | <pre> |
| + | x |
| + | o |
| + | | |
| + | O |
| + | </pre> |
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| The conjunction <math>x \land y</math> is written <math>x y.\!</math> | | The conjunction <math>x \land y</math> is written <math>x y.\!</math> |
| + | |
| + | This corresponds to the logical graph: |
| + | |
| + | <pre> |
| + | x y |
| + | O |
| + | </pre> |
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| The conjunction <math>x \land y \land z</math> is written <math>x y z.\!</math> | | The conjunction <math>x \land y \land z</math> is written <math>x y z.\!</math> |
| + | |
| + | This corresponds to the logical graph: |
| + | |
| + | <pre> |
| + | xyz |
| + | O |
| + | </pre> |
| | | |
| Etc. | | Etc. |
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| </pre> | | </pre> |
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− | The disjunction "''x'' or ''y'' or ''z''" is written "((''x'')(''y'')(''z''))". | + | The disjunction <math>x \lor y \lor z</math> is written <math>((x)(y)(z)).\!</math> |
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| This corresponds to the logical graph: | | This corresponds to the logical graph: |
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| Etc. | | Etc. |
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− | The implication "''x'' ⇒ ''y''" is written "(''x'' (''y'')), which can be read "not ''x'' without ''y''" if that helps to remember the form of expression. | + | The implication <math>x \Rightarrow y</math> is written <math>(x (y)),\!</math> which can be read "not <math>x\!</math> without <math>y\!</math>" if that helps to remember the form of expression. |
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| This corresponds to the logical graph: | | This corresponds to the logical graph: |
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| </pre> | | </pre> |
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− | Thus, the equivalence "''x'' ⇔ ''y''" has to be written somewhat inefficiently as a conjunction of to and fro implications: "(''x'' (''y''))(''y'' (''x''))". | + | Thus, the equivalence <math>x \Leftrightarrow y</math> has to be written somewhat inefficiently as a conjunction of two implications: <math>(x (y)) (y (x)).\!</math> |
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| This corresponds to the logical graph: | | This corresponds to the logical graph: |
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| ( (p (q)) (q (p)) ) = (p ( (q) )) ((p)(q)) | | ( (p (q)) (q (p)) ) = (p ( (q) )) ((p)(q)) |
| </pre> | | </pre> |
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− | No kidding …
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