I am busy having a war with Tarski's world but I'm obviously not winning right now.
I have the following sentence
∀x ∀y ∀z [(Cube(x) ∧ Cube(y) ∧ Cube(z)) → (x=y ∨ x=z ∨ y=z)]
On my world I have 3 small cubes as depicted below:
So now while arguing with the machine the sentence is resolving to False but what I am not understanding is why. We reach a point where it asks me if a = b is true and I argue yes but it insists false.
The question I am actually supposed to answer in conjunction with this sentence is to pick the correct option out of the below
- There are at most two cubes
- There are at most three cubes
- There are at least two cubes
- There are at least three cubes
Because of the meta-variables used here I am going with option 2.
I would think I am correct in this instance but I may stand corrected. Also, ultimately, why is it that this statement is ultimately outputting a false with 3 cubes of the same size in this world?
I have used the truth table approach and I am ultimately getting out a T for the antecedent of the conditional statement and a T for the consequent. With this I know that TT resolves to T under it's main connective in the conditional table and only F when TF.
The machine cannot be wrong, so where am I going wrong that I don't quite understand why ~(a = b)?
Many thanks for taking the time.
It means there are at most 2 cubes. It says, "For any cubes $x,y,z$, one of them is equal to another." This is false in models/worlds where there are three or more cubes: just take $x,y,z$ to be distinct cubes. In models where there are at most two cubes, then whichever cubes $x,y,z$ are, some pair of them have to be equal (it's the pigeonhole principle).