Truth Trees in QL: Example 1
We will test the following argument for validity:
| (x)Fx & (∃x)Gx |
| (x)(Gx ⊃ Hx) |
| (∃x)Hx |
We begin our tree with the premises and the negation of the conclusion. The reason why? It's the same as in PL. Every truth-tree is a test for consistency. If the premises and the negation of the conclusion are inconsistent, then the argument is valid.
| 1. | (x)Fx & (∃x)Gx | checked | conjunction. Used rule for conjunctions. | |||
| 2. | (x)(Gx ⊃ Hx) | |||||
| 3. | ~(∃x)Hx | checked | negation of quantified statement. Used quantifier negation. | |||
| 4. | (x)~Hx | |||||
| 5. | (x)Fx | left oonjunct of 1. | ||||
| 6. | (∃x)Gx | checked | Right conjunct of 1 | |||
| 7. | Ga | Instantiated 6. | ||||
| 8. | Fa | Instantiated 5. | ||||
| 9. | ~Ha | Instantiated 4 | ||||
| 10. | Ga ⊃ Ha | checked | conditional - use branching rule for conditionals | |||
| / | \ | |||||
| 11. | ~Ga | Ha | ||||
| X | X |