Prove that $\forall a \forall b \forall c ((a > b) \to (a + c > b + c))$ in an ordered field

Question

I want to show that this sentence holds, given the following axioms for an ordered field. I adopt the usual axioms of the field, plus the following:

1. $(a \geq b ~\wedge~b \geq a) \rightarrow a = b$
2. $(a \geq b ~\wedge~ b \geq c) \rightarrow a \geq c$
3. $a \geq b \vee b \geq a$
4. $a \geq b \rightarrow a + c \geq b + c$
5. $a \geq 0 \wedge b \geq 0 \rightarrow ab \geq 0$

I define $a > b$ as:

$a > b \equiv a \geq b \wedge \neg(a = b)$

Can anyone show me how to prove this using natural deduction? I'm guessing axiom 4. is the one I need to use somehow, but not matter how I rearrange it I can't seem to get close.

Answer 1

What you need is that $a>b$ is the same as $\neg b\ge a$. Then it follows from axiom 4 via contraposition and equation solving. Let $x = a+c$, and $y=b+c$:

$$a > b \Rightarrow x+(-c) > y+(-c) \Rightarrow x > y \Leftrightarrow a+c > b+c$$

But from your definition you have $a\ge b$ implies $a\ge b$ and if we assume that $b\ge a$ we would have $a=b$ which contradicts your definition. So by RAA we have $a > b \rightarrow \neg(b\ge a)$.

Now assume that $\neg (b\ge a)$ then we have by axiom 3 that $a \ge b$, we also have that $a\ne b$(*) so we conclude that $\neg(b\ge a) \Rightarrow a>b$ (which together with the previous paragraph shows equivalence).

(*) Normally one have $a\ge a$ as an axiom, but that follows directly from axiom 3 as it especially says $a\ge a\lor a\ge a$ which is logically the same as $a\ge a$. This can also be expressed as $a=b \Rightarrow b\ge a$. By conrapositive this means that $\neg (b\ge a) \Rightarrow a\ne b$.

Answer 2

If $a > b$ then as per definition $a ≠ b$ and $a ≥ b$ Now $a ≥ b \Rightarrow a + c ≥ b + c$ Now to prove $a + c > b + c$ (strictly) You just need to show that $a + c ≠ b +c$.. If it is equal then $a + c = b + c$ implies $a = b$ (using right cancellation law of field ) Which is not the case as $a>b$ This contradiction proves that $a + c > b + c$