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Ministry of Image - Fanfiction Printing

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safe2285135 artist:jan663 oc1012887 oc only741308 oc:canni soda459 earth pony552802 pony1714656 galacon980 galacon 201332 g42126632 animated133475 captain obvious743 clothes680890 earth pony oc29449 fancy mathematics263 formal wear394 mascot1681 math1007 necktie13212 numbers1516 paper5073 solo1506334 suit9872

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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.