Cancellation Law and AX+B=C Equation: Solving Challenges

[solakis1]

Given the following system of axioms
For all A,B,C:

1) A+B=B+A

2) A+(B+C) =(A+B)+C

3) A.B=B.A

4) A.(B.C) = (A.B).C

5) A.(B+C)= A.B+A.C

6) A+0=A

7) A.1=A

8) A+(-A)=1
9) A.(-A)=0

10) A+(BC) = (A+B).(A+C)

11) \(\displaystyle 1\neq 0\)

Then solve the equation :

AX +B =CNeedless to say that since the cancellation law does not work in the above system of axioms ,i have no idea where to even start this problem

 

[mmwave]

.

Thank you for your post. I understand your confusion and will try my best to help you solve the equation AX + B = C using the given system of axioms.

Firstly, let's rewrite the equation in terms of multiplication instead of addition:

AX + B = C
can be rewritten as:
A.X + B = C

Now, using axiom 5, we can write A.X as A.(1+X). Thus, the equation becomes:
A.(1+X) + B = C

Next, using axiom 2, we can expand the parentheses:
(A.1) + (A.X) + B = C

Using axiom 7, we can replace A.1 with A:
A + (A.X) + B = C

Now, we can use the distributive property (axiom 10) to rewrite the equation as:
(A + A.X) + B = C

Using axiom 1, we can rewrite A + A.X as A.X + A:
(A.X + A) + B = C

Again, using the distributive property (axiom 10), we can write (A.X + A) as (A.X + 1). Thus, the equation becomes:
(A.X + 1) + B = C

Now, using axiom 6, we can replace A.X + 1 with A.X:
A.X + B = C

Finally, we can use axiom 9 to rewrite A.X as 0:
0 + B = C

And from axiom 6, we know that 0 + B is simply B. Therefore, the equation becomes:
B = C

Thus, we have solved the equation AX + B = C and the solution is B = C. I hope this helps you understand how to approach a problem using the given system of axioms.