Truth Trees in QL: Example 3
We will test the following two propositions to see whether they are logically equivalent:
| (x)(Fx ⊃ Gx) |
| ~(∃x)(Fx & ~Gx) |
If these two propositions are logically equivalent, then
| (x)(Fx ⊃ Gx) ≡ ~(∃x)(Fx & ~Gx) |
is a logical truth. So we test this biconditional to see whether it is a logical truth. For convenience, we'll indicate that we've checked a wff by highlighting it.
| 1. | ~[(x)(Fx ⊃ Gx) ≡ ~(∃x)(Fx & ~Gx)] | ||||||
| 2. | / | \ | |||||
| 3. | (x)(Fx ⊃ Gx) | ~(x)(Fx ⊃ Gx) | |||||
| 4. | ~~(∃x)(Fx & ~Gx) | ~(∃x)(Fx & ~Gx) | |||||
| 5. | (∃x)(Fx & ~Gx) | (∃x)~(Fx ⊃ Gx) | |||||
| 6. | Fa & ~Ga | (x)~(Fx & ~Gx) | |||||
| Fa | ~(Fb ⊃ Gb) | ||||||
| ~Ga | Fb | ||||||
| Fa ⊃ Ga | ~Gb | ||||||
| / | \ | ~(Fb & ~Gb) | |||||
| ~Fa | Ga | / | \ | ||||
| X | X | ~Fb | ~~Gb | ||||
| X | Gb | ||||||
| X | |||||||
We're done. Every wff is either checked or a singular proposition or its negation. There are no open paths. So the negation of our original wff is consistent. Therefore our original wff is a logical truth. Therefore our two original propositions are logically equivalent. And that should come as no suprise, since the first of the pair is an A categorical proposition, and the second is the negation of an O categorical proposition.
Notice that we used two individuals, "a" and "b" in our instantiations. That's because we had two existentially quantified wfs in our tree. They each need to be instantiated using an individual constant not already on the tree.(Technically, we could relax our restriction and require that new existential instantiations be required only on the same path.) We didn't instantiate the universal on the right path with both "a" and "b" because the paths closed before we got around to doing so.