By J.Barkley Rosser

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**Example text**

8) Oui procedure to prove that some specific statement X , is not derivable in set theory is to establish two results: (1) Every statement derivable in set theory has the Boolean value 1. (2) The statement X,, does not have the Boolean value 1. In the present chapter we establish the result (1). X then kX. =X and 1 Y, then CZ. We will follow the formulation of Rosser [7], in which the sole rule is modus ponens. So the application of the rules of set theory is taken care of by the following theorem.

Let P E A and Q i E A for i E I. 68) Then there is a basis set B and an index i such that BcPhQi. 69) Q . 2, there is E P. 70) since it contains x . 72) C. 25. 73) Qi # 0. 6) there is a y with y E B, and y E Qi . 2, there is a basis set B, with y E B, and B, c Q i . 2, there is a basis set B with y E B and B E B, nB 2 . 69). C. Automorphisms In some cases, automorphisms of the Boolean algebra play an important role. 24. A mapping G is an automorphism of the Boolean algebra A iff it is one-to-one, its range and domain are both A, and for P, Q E A we have G(P A G(P v Q>= G(P) A G(Q) Q ) = G(P) v G ( Q ) G(P') = (G(P))'.

N(Y&Z)I XI. Y I Z i f a n d o n l y i f -(Z&X)). llXIl~llYllI~IZl1. By ( 3 . 37), k X = . YIZ iff IIXII’v(IIYIJ’vllZII)=l. 37) l l ~ l l ’ ~ ~ l IlZll)=(llXll l ~ l l ’ ~ A IIYII)*IIZII. 48). We now consider the axiom (x) . F(x) =iG(x) : 3 : (x)F(x) 3 (x)G(x). 10) and 3. 35) IIF(b) 3 G(b)II A IIF(b)II = IIF@)IIA IIW)II I IIG(b)ll. 9). The generalization to the case where F(x) and C(x) may contain other free variables besides x is easily handled, as is the case where one prefixes universal quantifiers to the axiom.