x/0=0 is a relatively tame axiom, it basically saying: project all zero reciprocals to zero. I think it might had something to do with projective geometry
-a/b = -(a/b) thus has to result from that since you are basically do the same as multiplying both sides by zero
It's only when you have nonzero y and then axiomise x/0=y for some pairs of (x,y) that will get you in trouble if you do not throw away associativity
y/1+1/0 = y/1+0 = y; (yx0+1x1)/(1x0) = (0+1)/0 =1/0=0
y/1+0=y/1; y/1+(0z)/z=(yxz+((0z)1+1))/(1x1)=(yz+1)/1=y+1
y/1+0=y/1; y/1+(0z)/z=(yxz+((0z)1+1))/(1xz)=(yz+1)/z=y+1/z
y/1+a=y/1; y/1+(az)/z=(yxz+((az)1+1)/(1xz)=(yz+az1+1)/z=y+a1+1/z
y/1+1/0 = y/1+0 = y; (yx0+1x1)/(1x0) = (0+1)/0 =1/0=0
y/1+0 = y/1+(0z)/z = y; (yxz+(0z)x1)/(1x0) = (yz+0z)/0 =yz/0+0z/0=0+0=0
y/1+0 = y/1+(0z)/z = y; (yxz+(0z)x1)/(1xz) = (yz+0z)/z =yz/z+0z/z=y+0=y
(y/1+1/0)(1x0)=yx0+1x1; y0/1+0/0=y0+0/0=y0
@LeakyNun Your issue is because 0/0=/=1 ,while for all nonzero x/x=1, thus you have a discontinuity there
Thus (x/0)0=00; x0/0=0; 0/0=0
This is why the product of reciprocal axiom fails for the zero reciprocals, leading to the formation of a new theorem y=y+1
since 0/0 should be 1, but it cannot, thus the only way for the system to adapt, is to have y=y+1
This is common phenomenon in many zero term algebra, that you end up with the theorem y=y+1
Protip: Whenever one meddles with some forms of division by zero, always write out the zero terms in full to see what's going on
So what you have here is not only that a/b+c/d=(ad+bc)/(bd) fails, but because of y=y+1 (and similarly y=1+y) for all y
and that means you have 1=0, thus you collapse into the trivial ring
In order to save this structure, you might want to figure out how to nuke 0/0 from the expansion by modifying the sum of fraction axiom
Alternately, you can discard one of the distributive laws and thus end up with two one sided identities 0 and 1
or, you can modify the distributive laws in the same way as wheels to deal with the occurrence of zero terms