Table 13
| Connective | Antecedent Switch | Set at | Consequent Switch | Set at |
|---|---|---|---|---|
| k₁ | A₁ | V | C₁ | k₂ |
| k₂ | A₂ | V | C₂ | V |
| k₃ | A₃ | k₁ | C₃ | k₅ |
| k₄ | A₄ | V | C₄ | V |
| k₅ | A₅ | k₄ | C₅ | k₆ |
| k₆ | A₆ | V | C₆ | V |
| k₇ | A₇ | k₃ | C₇ | V |
In any problem, statements that are different are numbered one after another 1, 2, 3, 4 ···. A statement that is repeated bears always the same number. In nearly all cases that are interesting, there will be repetitions of the statements. If any statement appeared with a “not” in it, we would turn up the denial switch for that statement (Unit 2).
The different connectives available on the machine are “and,” “or,” “if ··· then,” “if and only if.” If a “not” affected the compound statement produced by any connective, we would turn up the denial switch for that connective (Unit 8).
The last step in putting the problem on the machine is to connect the main connective of the whole compound statement to the yellow light output (Unit 12). In this problem the last “if-then,” k₇, is the main connective, the one that produces the whole compound statement. So we turn Stop Switch 7 (in Unit 7) that belongs to k₇ into the up position. There are a few more things to do, naturally, but the essential part of putting the information of the problem into the machine has now been described.