I am trying to understand how some basic properties of the real numbers can be proved from axioms expressed in predicate logic.
I start by accepting the field axioms of real numbers, in addition to the following axioms
- $(\forall x,y \in \mathbb R)((x \geq y \wedge y \geq x) \rightarrow x = y)$
- $(\forall x,y \in \mathbb R)((x \geq y \wedge y \geq z) \rightarrow x \geq z)$
- $(\forall x,y \in \mathbb R)(x \geq y \vee y \geq x)$
- $(\forall x,y \in \mathbb R)(x \geq y \rightarrow x + z \geq y + z)$
- $(\forall x,y \in \mathbb R)((x \geq 0 \wedge y \geq 0) \rightarrow xy \geq 0)$
As a first step, I want to prove that $(x = y) \rightarrow (x + z = y + z)$ using only appropriate manipulations of the axioms above using the laws of predicate logic. It is obviously intuitive to me that this is true, but not entirely obvious how to prove this statement formally.
I start by redefining these statements (leaving out quantifiers for simplicity) in terms of predicates P and Q, as follows:
Let $P(x,y) \equiv x \geq y $ and $Q(x,y) \equiv x = y$
Then, I reformulate these axioms as:
- $P(x,y) \wedge P(y,x) \rightarrow Q(x,y)$
- $P(x,y) \wedge P(y,z) \rightarrow P(x,z)$
- $P(x,y) \vee P(y,x)$
- $P(x,y) \rightarrow P(x + z,y + z)$
- $P(x,0) \wedge P(y, 0) \rightarrow P(xy,0)$
I need to prove that $Q(x,y) \rightarrow Q(x+z,y+z)$. I know that I can do this if I can prove that $P(x,y) \wedge P(y,x) \leftrightarrow Q(x,y)$. It is obvious to me that this is true.
I started by trying to prove that $Q(x,y) \rightarrow P(x,y) \wedge P(y,x)$. Once I have this I can prove the equivalence by combining this with 1. I think I need to show somehow that $P(x,y) \vee P(y,x)$ is not sufficient for $Q(x,y)$, because it is my understanding that 1. states that for them both to be true is sufficient for $Q(x,y)$ to be true, but there maybe other cases that make $Q(x,y)$ true. Can anyone help me out here?