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Assuming the associativity of multiplication among 2-adic relatives, we may compute the product <math>~\mathrm{m,}\mathrm{b,}\mathrm{r}~</math> by a brute force method as follows:

Assuming the associativity of multiplication among 2-adic relatives, we may compute the product <math>~\mathrm{m},\mathrm{b},\mathrm{r}~</math> by a brute force method as follows:

{| align="center" cellspacing="6" width="90%"

{| align="center" cellspacing="6" width="90%"

|

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<math>\begin{array}{lll}

<math>\begin{array}{lll}

\mathrm{m,}\mathrm{b,}\mathrm{r}

\mathrm{m},\mathrm{b},\mathrm{r}

& = &

& = &

(\mathrm{C}\!:\!\mathrm{C} ~+\!\!,~ \mathrm{I}\!:\!\mathrm{I} ~+\!\!,~ \mathrm{J}\!:\!\mathrm{J} ~+\!\!,~ \mathrm{O}\!:\!\mathrm{O})(\mathrm{O}\!:\!\mathrm{O})(\mathrm{D} ~+\!\!,~ \mathrm{O})

(\mathrm{C}\!:\!\mathrm{C} ~+\!\!,~ \mathrm{I}\!:\!\mathrm{I} ~+\!\!,~ \mathrm{J}\!:\!\mathrm{J} ~+\!\!,~ \mathrm{O}\!:\!\mathrm{O})(\mathrm{O}\!:\!\mathrm{O})(\mathrm{D} ~+\!\!,~ \mathrm{O})

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This avers that a man that is black that is rich is Othello, which is true on the premisses of our universe of discourse.

This says that a man that is black that is rich is Othello, which is true on the premisses of our present universe of discourse.

The stock associations of <math>\mathfrak{g}\mathit{o}\mathrm{h}</math> lead us to multiply out the product <math>~\mathrm{m},\!,\mathrm{b},\mathrm{r}~</math> along the following lines, where the trinomials of the form <math>(X\!:\!Y\!:\!Z)(Y\!:\!Z)(Z)\!</math> are the only ones that produce any non-null result, specifically, of the form <math>(X\!:\!Y\!:\!Z)(Y\!:\!Z)(Z) = X.\!</math>

Following the standard associative combinations of <math>\mathfrak{g}\mathit{o}\mathrm{h},</math> the product <math>~\mathrm{m},\!,\mathrm{b},\mathrm{r}~</math> is multiplied out along the following lines, where the trinomials of the form <math>(X\!:\!Y\!:\!Z)(Y\!:\!Z)(Z)\!</math> are the only ones that produce a non-null result, namely, <math>(X\!:\!Y\!:\!Z)(Y\!:\!Z)(Z) = X.\!</math>

:{| cellpadding="4"

{| align="center" cellspacing="6" width="90%"

| m,,b,r

|

| =

| (C:C:C +, I:I:I +, J:J:J +, O:O:O)(O:O)(D +, O)

\mathrm{m},\!,\mathrm{b},\mathrm{r}

& = &

| &nbsp;

(\mathrm{C}\!:\!\mathrm{C}\!:\!\mathrm{C} ~+\!\!,~ \mathrm{I}\!:\!\mathrm{I}\!:\!\mathrm{I} ~+\!\!,~ \mathrm{J}\!:\!\mathrm{J}\!:\!\mathrm{J} ~+\!\!,~ \mathrm{O}\!:\!\mathrm{O}\!:\!\mathrm{O})(\mathrm{O}\!:\!\mathrm{O})(\mathrm{D} ~+\!\!,~ \mathrm{O})

| =

| (O:O:O)(O:O)(O)

& = &

(\mathrm{O}\!:\!\mathrm{O}\!:\!\mathrm{O})(\mathrm{O}\!:\!\mathrm{O})(\mathrm{O})

| &nbsp;

| =

& = &

| O

\mathrm{O}

|}

|}

So we have that m,,b,r = m,b,r.

So we have that <math>\mathrm{m},\!,\mathrm{b},\mathrm{r} ~=~ \mathrm{m},\mathrm{b},\mathrm{r}.</math>

In closing, observe that the teridentity relation has turned up again in this context, as the second comma-ing of the universal term itself:

In closing, observe that the teridentity relation has turned up again in this context, as the second comma-ing of the universal term itself: