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A truth table for this problem is shown below. Each of the first four columns represents one of the elements of the Herbrand base for this language. The two middle columns represent our premises, and the final column represents the conclusion.
| p(a) |
p(b) |
q(a) |
q(b) |
p(a) ∨ p(b) |
∀x.(p(x) ⇒ q(x)) |
∃x.q(x) |
| 1 |
1 |
1 |
1 |
1 |
1 |
1 |
| 1 |
1 |
1 |
0 |
1 |
0 |
1 |
| 1 |
1 |
0 |
1 |
1 |
0 |
1 |
| 1 |
1 |
0 |
0 |
1 |
0 |
0 |
| 1 |
0 |
1 |
1 |
1 |
1 |
1 |
| 1 |
0 |
1 |
0 |
1 |
1 |
1 |
| 1 |
0 |
0 |
1 |
1 |
0 |
1 |
| 1 |
0 |
0 |
0 |
1 |
0 |
0 |
| 0 |
1 |
1 |
1 |
1 |
1 |
1 |
| 0 |
1 |
1 |
0 |
1 |
0 |
1 |
| 0 |
1 |
0 |
1 |
1 |
1 |
1 |
| 0 |
1 |
0 |
0 |
1 |
0 |
0 |
| 0 |
0 |
1 |
1 |
0 |
1 |
1 |
| 0 |
0 |
1 |
0 |
0 |
1 |
1 |
| 0 |
0 |
0 |
1 |
0 |
1 |
1 |
| 0 |
0 |
0 |
0 |
0 |
1 |
0 |
Looking at the table, we see that there are 12 truth assignments that make the first premise true and nine that make the second premise true and five that make them both true (rows 1, 5, 6, 9, and 11). Note that every truth assignment that makes both premises true also makes the conclusion true. Hence, the premises logically entail the conclusion.
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