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safe2229740 artist:jan661 oc976585 oc only713155 oc:canni soda369 earth pony528605 pony1659788 galacon601 galacon 201331 g42091837 animated129527 captain obvious728 clothes659895 earth pony oc27633 fancy mathematics264 formal wear364 mascot1644 math994 necktie12221 numbers1483 paper4857 solo1474235 suit9531

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Blacklight Shining

@B2C  
For any numbers x and y, x > y only if there exists a number such that y + that number = x.
 
x, y := 6, 9  
y + (-3) = x  
x > y  
Oops?
 
You forgot to say that that number must be greater than 0.
CocoaNut

Silly pony. You need about 600 more steps to prove that 2 > 1.
 
And first you need to prove that 1 + 1 even equals 2 if you’re going to take that route.
B2C
The End wasn't The End - Found a new home after the great exodus of 2012

We now that 1 is s(0) and that 2 is s(s(0)).  
From the definition of addition s(0) + s(0) = s(0 + s(0)). Axiomatically, for all x, x + 0 = x. From this, 0 + 0 = 0, and if 0 + x = x, then 0 + s(x) = s(0 + x) = s(x), and so 0 + x = 0 for all x. Thus, 1 + 1 = s(0) + s(0) = s(0 + s(0)) = s(s(0)) = 2. For any numbers x and y, x > y only if there exists a number such that y + that number = x. Since 1 + 1 = 2, 2 > 1.