Section 1.16

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"Proof: ."

"Hence 1.14(b) implies that ."

As shown above, .

"Then (a) gives ."

"The first equality in (c) comes from , combined with 1.14(c)."

Then , and by 1.14(c), .

"The other half of (c) is proved in the same way."

Then , and by 1.14(c), .

"Finally,  by (c) and 1.14(d)."

Section 1.15

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"The proof is so similar to that of Proposition 1.14 that we omit it."

(a) 

        

(b)

therefore, by (a), .

(c)

Applying (a) gives .

(d)  If, for some , we have , then by (c) we have .  Then substitution gives  and we have the following.

Then , and by (a), .

Section 1.14

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"Proof: If  , the axioms (A) give

."

Here is how each step is arrived at:

"Take  in (a) to obtain (c)."

so (a) gives .

"Since , (c) (with  in place of ) gives (d)."

If, for some , we have , then by (c), .  Then substitution gives .  Then we have:

and by (a), .

Section 1.11

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"It follows that  for every ."

This is because all  are the upper bounds of .   is the least upper bounds, i.e. the least of all the .  Therefore  for every .