Of the 16 units 13 are input units. They control the setup of the machine so that it can solve a problem. Of the 13 input units, those that have the most to do with taking in the problem are shown in [Table 12].
Table 12
| Unit | Name of Switches | Mark | Kind of Switch | Switch Settings |
|---|---|---|---|---|
| 3 | Statement | V₁ to V₁₂ | Dial | Statements 1 to 12 or constant T or F |
| 2 | Statement denial | ~ | Snap | Affirmative (down) or negative (up) |
| 4 | Connective | k₁ to k₁₁ | Dial | ∧ (and), ∨ (or), ▲ (if-then), ▼ (if and only if) |
| 8 | Connective denial | ~ | Snap | Affirmative (down) or negative (up) |
| 5 | Antecedent | A₁ to A₁₁ | Dial | V or various k’s |
| 6 | Consequent | C₁ to C₁₁ | Dial | V or various k’s |
| 7 | Stop | S₁ to S₁₁ | Snap | Not connected (down) or connected (up) |
The first step in putting a problem on the machine is to express the whole problem as a single compound statement that we want to know the truth or falsity of. We express the single compound statement in a form such as the following:
V k V k V k V k V k V k V k V k V k V k V k V
where each V represents a statement, each k represents a connective, and we know the grouping, or in other words, we know the antecedent and consequent of each connective.
For example, let us choose a problem with an obvious answer:
Problem. Given: statement 1 is true; and if statement 1 is true, then statement 2 is true; and if statement 2 is true, then statement 3 is true; and if statement 3 is true, then statement 4 is true. Is statement 4 true?
How do we express this whole problem in a form that will go on the machine? We express the whole problem as a single compound statement that we want to know the truth or falsity of:
If [1 and (if 1 then 2) and (if 2 then 3) and (if 3 then 4)], then 4