Line 1,191: |
Line 1,191: |
| ===Computation Summary : <math>f(u, v) = \texttt{((u)(v))}</math>=== | | ===Computation Summary : <math>f(u, v) = \texttt{((u)(v))}</math>=== |
| | | |
− | <pre>
| + | Figure 1.1 shows the expansion of <math>f = \texttt{((u)(v))}</math> over <math>[u, v]\!</math> to produce the expression: |
− | Figure 1.1 expands f = ((u)(v)) over [u, v] to produce | + | |
− | the equivalent exclusive disjunction uv + u(v) + (u)v.
| + | {| align="center" cellpadding="8" width="90%" |
| + | | <math>\texttt{uv ~+~ u(v) ~+~ (u)v}</math> |
| + | |} |
| + | |
| + | Figure 1.2 shows the expansion of <math>\operatorname{E}f = \texttt{((u + du)(v + dv))}</math> over <math>[u, v]\!</math> to produce the expression: |
| | | |
− | Figure 1.2 expands Ef = ((u + du)(v + dv)) over [u, v] to arrive at
| + | {| align="center" cellpadding="8" width="90%" |
− | Ef = uv (du dv) + u(v) (du (dv)) + (u)v ((du) dv) + (u)(v)((du)(dv)).
| + | | <math>\texttt{uv~(du~dv) ~+~ u(v)~(du (dv)) ~+~ (u)v~((du) dv) ~+~ (u)(v)~((du)(dv))}</math> |
| + | |} |
| | | |
| + | <pre> |
| Ef tells you what you would have to do, from where you are in the | | Ef tells you what you would have to do, from where you are in the |
| universe [u, v], if you want to end up in a place where f is true. | | universe [u, v], if you want to end up in a place where f is true. |