Addition +
Define addition + : Z2 → Z as follows:
∀x y ∈ Z x+y := x – Neg(y)
It follows that:
THEOREM 4 ∀x y ∈ Z x-y = x + Neg(y)
PROOF
x + Neg(y) = x – Neg(Neg(y)) = x-y
END of PROOF
Let x=y in the above theorem. Then we get:
∀x ∈Z x + Neg(x) = 0
THEOREM 5 ∀x y ∈ Z x+y = y+x
PROOF
x+y = x – Neg(y) = x – (0 – y) = y – (0 – x) = y – Neg(x) = y+x
END of PROOF
THEOREM 6 ∀x y ∈ Z Neg(x-y) = Neg(x) – Neg(y)
Neg(x-y) = y-x = y + Neg(x) = Neg(x) + y = Neg(x) – Neg(y)
END of PROOF
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THEOREM 7 ∀x y ∈ Z Neg(x+y) = Neg(x) + Neg(y)
PROOF
Neg(x+y) = Neg(x – Neg(y))
= Neg(x) – Neg(Neg(y)) = Neg(x) + Neg(y)
END of PROOF
THEOREM 8 ∀x y z∈ Z (x-y)-z = (x-z)-y = x – (y+z)
PROOF
x – (y+z) = x – (y – Neg(z))
= Neg(z) – (y – x) = Neg(z) – Neg(x-y)
= (x-y) + Neg(z) = (x-y) – Neg(Neg(z)) = (x-y)-z
Now the rest is obvious. END of PROOF
THEOREM 9 ∀x y z ∈ Z (x+y)+z = x+(y+z)
PROOF
(a+b)+c = (a – Neg(b)) – Neg(c) = a – ((Neg(b) + Neg(c))
= a – Neg(b+c) = a + (b + c)
END of PROOF