Bourbaki

Exercise 8, Appendix 0

Tag 03H4, permanent. Cite it as [Bourbaki, Tag 03H4].

Theory of Sets, I, 63

Bourbaki marks this one as harder than the rest.

Let R1R_1, R2R_2, ..., RnR_n be the logically irreducible relations in a theory T\mathscr{T} (Exercise 6). Assign to each of the relations RiR_i one of the signs 0,1,20, 1, 2. To each relation in a logical construction with base R1R_1, R2R_2, ..., RnR_n we assign one of the signs 0, 1, 2 in accordance with the following symbolical rules (Exercise 7) :

¬0=1,¬1=0,¬2=2;\neg 0 = 1, \qquad \neg 1 = 0, \qquad \neg 2 = 2;
∨00=∨01=∨02=∨10=∨20=∨22=0;\vee 00 = \vee 01 = \vee 02 = \vee 10 = \vee 20 = \vee 22 = 0;
∨11=1,∨12=∨21=2.\vee 11 = 1, \qquad \vee 12 = \vee 21 = 2.

(a) If RR is logically constructed from R1R_1, R2R_2, ..., RnR_n, the sign carried by RR is independent of the logical construction, with base R1R_1, R2R_2, ..., RnR_n, in which RR appears.

(b) Suppose that the only axioms of T\mathscr{T} are those provided by the schemes S2, S3, S4. Let RR be a theorem in T\mathscr{T}. Show that however the signs 0, 1, 2 are assigned to the logical components of RR, the corresponding sign carried by RR is 0. On the other hand, if SS is logically irreducible and has sign 2, then the sign carried by (S(S or S)⇒SS) \Rightarrow S is 2. Deduce that T\mathscr{T} is not equivalent to the theory which has the same signs as T\mathscr{T} and whose axioms are provided by the schemes S1, S2, S3, S4.

(c) Establish an analogous result for theories based only on the schemes S1, S3, S4, or on the schemes S1, S2, S4. (Use respectively the following rules :

¬0=1,¬1=0,¬2=2,∨00=∨01=∨10=∨02=∨20=0,\neg 0 = 1, \quad \neg 1 = 0, \quad \neg 2 = 2, \quad \vee 00 = \vee 01 = \vee 10 = \vee 02 = \vee 20 = 0,
∨11=1,∨12=∨21=1,∨22=1;\vee 11 = 1, \qquad \vee 12 = \vee 21 = 1, \qquad \vee 22 = 1;
¬0=1,¬1=2,¬2=0,\neg 0 = 1, \qquad \neg 1 = 2, \qquad \neg 2 = 0,
∨00=∨01=∨10=∨02=∨20=∨21=0,\vee 00 = \vee 01 = \vee 10 = \vee 02 = \vee 20 = \vee 21 = 0,
∨11=∨12=1,∨22=2.)\vee 11 = \vee 12 = 1, \qquad \vee 22 = 2.)

(d) Establish an analogous result for a theory based only on the schemes S1, S2, S3. (Use four signs 0, 1, 2, 3 and the following rules :

¬0=1,¬1=0,¬2=3,¬3=0,\neg 0 = 1, \qquad \neg 1 = 0, \qquad \neg 2 = 3, \qquad \neg 3 = 0,
∨00=∨01=∨10=∨02=∨20=∨03=∨30=∨23=∨32=0,\vee 00 = \vee 01 = \vee 10 = \vee 02 = \vee 20 = \vee 03 = \vee 30 = \vee 23 = \vee 32 = 0,
∨11=1,∨12=∨21=∨22=2,∨13=∨31=∨33=3.)\vee 11 = 1, \qquad \vee 12 = \vee 21 = \vee 22 = 2, \qquad \vee 13 = \vee 31 = \vee 33 = 3.)
Solution

No solution has been written yet.

Cited by

Nothing.

Cites

Transcribed from the printed volume. Not translated.