Changes

===Computation Summary for Logical Equality===

===Computation Summary for Logical Equality===

Figure&nbsp;2.1 shows the expansion of <math>g = \texttt{((u, v))}</math> over <math>[u, v]\!</math> to produce the expression:

Figure&nbsp;2.1 shows the expansion of <math>g = \texttt{((} u \texttt{,~} v \texttt{))}\!</math> over <math>[u, v]\!</math> to produce the expression:

{| align="center" cellpadding="8" width="90%"

{| align="center" cellpadding="8" width="90%"

| <math>\texttt{uv} ~+~ \texttt{(u)(v)}</math>

|

<math>\begin{matrix}

uv & + & \texttt{(} u \texttt{)(} v \texttt{)}

\end{matrix}</math>

|}

|}

Figure&nbsp;2.2 shows the expansion of <math>\mathrm{E}g = \texttt{((u + du, v + dv))}</math> over <math>[u, v]\!</math> to produce the expression:

Figure&nbsp;2.2 shows the expansion of <math>\mathrm{E}g = \texttt{((} u + \mathrm{d}u \texttt{,~} v + \mathrm{d}v \texttt{))}\!</math> over <math>[u, v]\!</math> to produce the expression:

{| align="center" cellpadding="8" width="90%"

{| align="center" cellpadding="8" width="90%"

| <math>\texttt{uv} \cdot \texttt{((du, dv))} + \texttt{u(v)} \cdot \texttt{(du, dv)} + \texttt{(u)v} \cdot \texttt{(du, dv)} + \texttt{(u)(v)} \cdot \texttt{((du, dv))}</math>

|

<math>\begin{matrix}

uv \cdot \texttt{((} \mathrm{d}u \texttt{,~} \mathrm{d}v \texttt{))} & + &

u \texttt{(} v \texttt{)} \cdot \texttt{(} \mathrm{d}u \texttt{,~} \mathrm{d}v \texttt{)} & + &

\texttt{(} u \texttt{)} v \cdot \texttt{(} \mathrm{d}u \texttt{,~} \mathrm{d}v \texttt{)} & + &

\texttt{(} u \texttt{)(} v \texttt{)} \cdot \texttt{((} \mathrm{d}u \texttt{,~} \mathrm{d}v \texttt{))}

\end{matrix}</math>

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|}

<math>\mathrm{E}g</math> tells you what you would have to do, from where you are in the universe <math>[u, v],\!</math> if you want to end up in a place where <math>g\!</math> is true.  In this case, where the prevailing proposition <math>g\!</math> is <math>\texttt{((u, v))},</math> the component <math>\texttt{uv} \cdot \texttt{((du, dv))}</math> of <math>\mathrm{E}g</math> tells you this:  If <math>u\!</math> and <math>v\!</math> are both true where you are, then change either both or neither of <math>u\!</math> and <math>v\!</math> at the same time, and you will attain a place where <math>\texttt{((du, dv))}</math> is true.

In general, <math>\mathrm{E}g\!</math> tells you what you would have to do, from wherever you are in the universe <math>[u, v],\!</math> if you want to end up in a place where <math>g\!</math> is true.  In this case, where the prevailing proposition <math>g\!</math> is <math>\texttt{((} u \texttt{,~} v \texttt{))},\!</math> the component <math>uv \cdot \texttt{((} \mathrm{d}u \texttt{,~} \mathrm{d}v \texttt{))}\!</math> of <math>\mathrm{E}g\!</math> tells you this:  If <math>u\!</math> and <math>v\!</math> are both true where you are, then change either both or neither of <math>u\!</math> and <math>v\!</math> at the same time, and you will attain a place where <math>\texttt{((} u \texttt{,~} v \texttt{))}\!</math> is true.

Figure&nbsp;2.3 shows the expansion of <math>\mathrm{D}g</math> over <math>[u, v]\!</math> to produce the expression:

Figure&nbsp;2.3 shows the expansion of <math>\mathrm{D}g\!</math> over <math>[u, v]\!</math> to produce the expression:

{| align="center" cellpadding="8" width="90%"

{| align="center" cellpadding="8" width="90%"

| <math>\texttt{uv} \cdot \texttt{(du, dv)} ~+~ \texttt{u(v)} \cdot \texttt{(du, dv)} ~+~ \texttt{(u)v} \cdot \texttt{(du, dv)} ~+~ \texttt{(u)(v)} \cdot \texttt{(du, dv)}</math>

|

<math>\begin{matrix}

uv \cdot \texttt{(} \mathrm{d}u \texttt{,~} \mathrm{d}v \texttt{)} & + &

u \texttt{(} v \texttt{)} \cdot \texttt{(} \mathrm{d}u \texttt{,~} \mathrm{d}v \texttt{)} & + &

\texttt{(} u \texttt{)} v \cdot \texttt{(} \mathrm{d}u \texttt{,~} \mathrm{d}v \texttt{)} & + &

\texttt{(} u \texttt{)(} v \texttt{)} \cdot \texttt{(} \mathrm{d}u \texttt{,~} \mathrm{d}v \texttt{)}

\end{matrix}</math>

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|}

<math>\mathrm{D}g</math> tells you what you would have to do, from where you are in the universe <math>[u, v],\!</math> if you want to bring about a change in the value of <math>g,\!</math> that is, if you want to get to a place where the value of <math>g\!</math> is different from what it is where you are.  In the present case, where the ruling proposition <math>g\!</math> is <math>\texttt{((u, v))},</math> the term <math>\texttt{uv} \cdot \texttt{(du, dv)}</math> of <math>\mathrm{D}g</math> tells you this:  If <math>u\!</math> and <math>v\!</math> are both true where you are, then you would have to change one or the other but not both <math>u\!</math> and <math>v\!</math> in order to reach a place where the value of <math>g\!</math> is different from what it is where you are.

In general, <math>\mathrm{D}g\!</math> tells you what you would have to do, from wherever you are in the universe <math>[u, v],\!</math> if you want to bring about a change in the value of <math>g,\!</math> that is, if you want to get to a place where the value of <math>g\!</math> is different from what it is where you are.  In the present case, where the ruling proposition <math>g\!</math> is <math>\texttt{((} u \texttt{,~} v \texttt{))},\!</math> the term <math>uv \cdot \texttt{(} \mathrm{d}u \texttt{,~} \mathrm{d}v \texttt{)}\!</math> of <math>\mathrm{D}g\!</math> tells you this:  If <math>u\!</math> and <math>v\!</math> are both true where you are, then you would have to change one or the other but not both <math>u\!</math> and <math>v\!</math> in order to reach a place where the value of <math>g\!</math> is different from what it is where you are.

Figure&nbsp;2.4 approximates <math>\mathrm{D}g</math> by the linear form <math>\mathrm{d}g</math> that expands over <math>[u, v]\!</math> as follows:

Figure&nbsp;2.4 approximates <math>\mathrm{D}g\!</math> by the linear form <math>{\mathrm{d}g}\!</math> that expands over <math>[u, v]\!</math> as follows:

{| align="center" cellpadding="8" width="90%"

{| align="center" cellpadding="8" width="90%"

|

|

<math>\begin{array}{lll}

<math>\begin{array}{*{9}{l}}

\mathrm{d}g

\mathrm{d}g

& = & \texttt{uv}\!\cdot\!\texttt{(du, dv)} + \texttt{u(v)}\!\cdot\!\texttt{(du, dv)} + \texttt{(u)v}\!\cdot\!\texttt{(du, dv)} + \texttt{(u)(v)}\!\cdot\!\texttt{(du, dv)}

& = & uv \cdot \texttt{(} \mathrm{d}u \texttt{,~} \mathrm{d}v \texttt{)}

\\ \\

& + & u \texttt{(} v \texttt{)} \cdot \texttt{(} \mathrm{d}u \texttt{,~} \mathrm{d}v \texttt{)}

& = & \texttt{(du, dv)}

& + & \texttt{(} u \texttt{)} v \cdot \texttt{(} \mathrm{d}u \texttt{,~} \mathrm{d}v \texttt{)}

& + & \texttt{(} u \texttt{)(} v \texttt{)} \cdot \texttt{(} \mathrm{d}u \texttt{,~} \mathrm{d}v \texttt{)}

\\[8pt]

& = & \texttt{(} \mathrm{d}u \texttt{,~} \mathrm{d}v \texttt{)}

\end{array}</math>

\end{array}</math>

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Figure&nbsp;2.5 shows what remains of the difference map <math>\mathrm{D}g</math> when the first order linear contribution <math>\mathrm{d}g</math> is removed, namely:

Figure&nbsp;2.5 shows what remains of the difference map <math>\mathrm{D}g\!</math> when the first order linear contribution <math>{\mathrm{d}g}\!</math> is removed, namely:

{| align="center" cellpadding="8" width="90%"

{| align="center" cellpadding="8" width="90%"

|

|

<math>\begin{matrix}

<math>\begin{array}{*{9}{l}}

\mathrm{r}g

\mathrm{r}g

& = & \texttt{uv} \cdot \texttt{0} & + & \texttt{u(v)} \cdot \texttt{0} & + & \texttt{(u)v} \cdot \texttt{0} & + & \texttt{(u)(v)} \cdot \texttt{0}

& = & uv \cdot 0

\\ \\

& + & u \texttt{(} v \texttt{)} \cdot 0

& = & \texttt{0}

& + & \texttt{(} u \texttt{)} v \cdot 0

\end{matrix}</math>

& + & \texttt{(} u \texttt{)(} v \texttt{)} \cdot 0

\\[8pt]

& = & 0

\end{array}</math>

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