Verifying proof for the following question I wanted to check my proof if it is correct or not. Using these field axioms:

(i)Trichotomy Property:  Exactly one of $x&lt;y$, $y&lt;x$, or $x=y$ hold.

(ii)Transitivity: if $x&lt;y$ and $y&lt;z$ (which we could write in shorthand as $x&lt;y&lt;z$), then $x&lt;z$.

(iii)If $x&lt;y$ then $x+z &lt; y+z$.

(iv) If $x&lt;y$ and $z&gt;0$ then $xz&lt;yz$.


  
  (A1) Addition is commutative
  (A2) Addition is associative
  (A3) Addition has a neutral element $0$
  (A4) Any element has an additive inverse
  (A5) Multiplication is commutative
  (A6) Multiplication is associative
  (A7) Multiplication has a neutral element $1$
  (A8) Any non-zero element has a multiplicative inverse
  (A9) Multiplication distributes over addition
  



Prove that for all $a,\, b,\, c\in\mathbb{R}$, if $a&lt;b$ and $c&lt;0$ then $bc &lt; ac$.


Since $a&lt;b$, by (A4), we have that $$
0 = a + (-a) &lt; b + (-a) = b - a.$$ 
Since $b - a &gt; 0$ and $c &lt; 0$, then $c(b-a) &lt; 0$. By (A9), this is equivalent to $cb-ca&lt;0$. By (A3), $$
cb=cb-ca+ca&lt;0+ca=ca.$$
Thus $cb&lt;ca$.
"Since b−a>0 and c&lt;0, then c(b−a)&lt;0"

Why do you state this. Either this is what you wish to prove, or (and I suspect this) you are using  iv) but not saying so.

So you should say so.

Since $(b-a)&gt; 0$ and $c &lt; 0$ then by iv) $c(b-a) &lt; 0\cdot (b-a)$ 

But notice. You don't have an axiom that $0\cdot x = 0$(!!!!)

You have to prove that.

Hint 1: $0d = (0+0)d$ 

Hint 2: $-0d$ must exist.

....

Once you prove that you have 

$c(b-a) &lt;0$

So by A9 $c(b-a) = c(b+(-a)) = cb +c(-a)$

So $cb + c(-a) &lt; 0$ and 

....

Now we need to prove $c(-a) = -ca$.

Hint: $c(-a) + ca = c(-a + a)$.

...

Now we have $cb - ca &lt;0$ and ... you can finish your proof.
Your proof has a gap.  How do you know that $b-a \gt 0 \land c \lt 0 \Rightarrow c(b-a) \lt 0$?  I'm not saying you're wrong -- I'm just saying you haven't yet proved it.

Edit to add:

Lemma:  If $c \lt 0 \land x \lt y, \text{ then } xc \gt yc$.

Proof:  $c+(-c) = 0 \land c \lt 0 \Rightarrow -c \gt 0$.  (Use property (iii), adding $-c$ to both sides of the inequality.)

Thus, $-xc \lt -yc$.  Add $xc + yc$ to both sides:  $yc  \lt xc$ and the Lemma is proved.