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MyWikiBiz, Author Your Legacy — Thursday November 28, 2024
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====Example 6====
 
====Example 6====
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We have now developed two ways of computing a logical involution that raises a 2-adic relative term to the power of a 1-adic absolute term, for example, <math>\mathit{l}^\mathrm{w}\!</math> for "lover of every woman".
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The discussion up to this point has developed two ways of computing a logical involution that raises a 2-adic relative term to the power of a 1-adic absolute term, for example, <math>\mathit{l}^\mathrm{w}\!</math> for "lover of every woman".
    
The first method operates in the medium of set theory, expressing the denotation of the term <math>\mathit{l}^\mathrm{w}\!</math> as the intersection of a set of relational applications:
 
The first method operates in the medium of set theory, expressing the denotation of the term <math>\mathit{l}^\mathrm{w}\!</math> as the intersection of a set of relational applications:
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| valign="top" | 5.
 
| valign="top" | 5.
| Compute the value <math>\mathfrak{L}_{ux}^{\mathfrak{W}_x} = (\mathfrak{L}_{ux}\!\Leftarrow\!\mathfrak{W}_x)</math> for each <math>x\!</math> in the middle row.
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| Compute the value <math>\mathfrak{L}_{ux}^{\mathfrak{W}_x} = (\mathfrak{L}_{ux} >\!\!\!-~ \mathfrak{W}_x)</math> for each <math>x\!</math> in the middle row.
 
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| valign="top" | 6.
 
| valign="top" | 6.
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As a general observation, then, we know that the value of <math>(\mathfrak{L}^\mathfrak{W})_u</math> goes to <math>0\!</math> just as soon as we find an <math>x \in X</math> such that <math>\mathfrak{L}_{ux} = 0</math> and <math>\mathfrak{W}_x = 1,</math> in other words, such that <math>(u, x) \notin L</math> but <math>u \in W.</math>  If there is no such <math>x\!</math> then <math>(\mathfrak{L}^\mathfrak{W})_u = 1.</math>
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As a general observation, we know that the value of <math>(\mathfrak{L}^\mathfrak{W})_u</math> goes to <math>0\!</math> just as soon as we find an <math>x \in X</math> such that <math>\mathfrak{L}_{ux} = 0</math> and <math>\mathfrak{W}_x = 1,</math> in other words, such that <math>(u, x) \notin L</math> but <math>u \in W.</math>  If there is no such <math>x,\!</math> then <math>(\mathfrak{L}^\mathfrak{W})_u = 1.</math>
    
===Commentary Note 12.2===
 
===Commentary Note 12.2===
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