By Wolfram Pohlers (author), Thomas Glaß (editor)
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A) If M j= G ! F and x 2= FV(M) FV(G), where FV(M) = fFV(H) : H 2 M g then M j= G ! 8xF b) If M j= F ! G and x 2= FV(M) FV(G), then M j= 9xF ! G: Proof. Again both claims are dual. Thus it su ces to prove b). Let S be an Lstructure and an S -assignment such that S j= M ]: Then we have S j= F ! G ]: If S j= G ] we are done. Thus assume S 6j= G ] which entails that If is any S -assignment such that S 6j= F ]: x , then we have S j= M ] i S j= M ] as well as S j= G ] i S j= G ] because x 2= FV(M fGg): Hence S j= M ] and S 6j= G ] which by M j= F !
Let TF be the theory of elds. TOF the theory of ordered elds is obtained by TF adding a binary predicate symbol < and the following axioms: 1. 8x(:x < x) 2. 8x8y8z(x < y ^ y < z ! x < z) 3. 8x8y(x < y _ x = y _ y < x) 4. 8x8y8z(x < y ! x + z < y + z) 5. 8x8y8z(x < y ^ 0 < z ! x z < y z) A structure S j= TOF is called Archimedian ordered of for any s 2 S there is an n 2 IN such that S j= x < 1| + :{z: : + 1} s] n-times a) Prove that there is no theory T such that for all LI-structures S (cf. 6) S j= T , S is an Archimedian ordered eld.
We mean by the cardinality of a language the cardinality of the set of its non-logical symbols. 5. 11. Let L be a rst order language. Then the Henkin extension LH has the cardinality card(LH ) = max(@0 card(L)): Proof. This is indicated in the exercises for card(L) @0 : But the general statement is obtained by a similar proof. 12. Let M be a set of L-sentences. e. it is card(S) = max(@0 card(L)): Proof. For card(L) @0 this will be proved in the exercises. Nearly the same proof yields the theorem.