Truth Trees in QL: Example 2

We will test the following proposition to see whether it is logically true:

(∃x)Gx  ⊃ (x)Gx

We begin our tree with the negation of the proposition. The reason why? It's the same as in PL. Every truth-tree is a test for consistency. If the negation of a wff is inconsistent, then the wff is a logical truth.

1.     ~[(∃x)Gx  ⊃ (x)Gx]   checked. negation of wff we are testing.
2.     (∃x)Gx   checked. negation of conditional rule
3.     ~(x)Gx   checked. negation of conditional rule
4.     (∃x)~Gx   checked. quantifier negation rule
5.     Ga     instantiate 2.
6.     ~Gb     instantiate 4.

We're done. Every wff is either checked or a singular proposition or its negation. There is an open path. So the negation of our original wff is not inconsistent. Therefore our original wff is not a logical truth.

Notice that we had to use 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.

It should be no surprise that our wff is not logically true. It says that if something has a property, then everything has that property.  Is there a universe in which this true? How many individuals does such a universe have? What our tree shows is that it is false in a universe of two individuals.