Volume 11, 1999

University of Bialystok

Copyright (c) 1999 Association of Mizar Users

### The abstract of the Mizar article:

### Predicate Calculus for Boolean Valued Functions. Part VI

**by****Shunichi Kobayashi**- Received October 19, 1999
- MML identifier: BVFUNC14

- [ Mizar article, MML identifier index ]

environ vocabulary EQREL_1, T_1TOPSP, PARTIT1, BOOLE, SETFAM_1, FUNCOP_1, RELAT_1, FUNCT_1, CANTOR_1, CAT_1, FUNCT_4, BVFUNC_2; notation TARSKI, XBOOLE_0, ENUMSET1, ZFMISC_1, SUBSET_1, RELAT_1, FUNCT_1, SETFAM_1, EQREL_1, CANTOR_1, CQC_LANG, PARTIT1, BVFUNC_1, BVFUNC_2, FUNCT_4; constructors DOMAIN_1, CANTOR_1, BVFUNC_2, BVFUNC_1, FUNCT_4; clusters PARTIT1, SUBSET_1, FUNCT_4, CQC_LANG, XBOOLE_0; requirements SUBSET, BOOLE; begin :: Chap. 1 Preliminaries reserve Y for non empty set, G for Subset of PARTITIONS(Y), A,B,C,D for a_partition of Y; theorem :: BVFUNC14:1 for z being Element of Y, PA,PB being a_partition of Y holds EqClass(z,PA '/\' PB) = EqClass(z,PA) /\ EqClass(z,PB); theorem :: BVFUNC14:2 G={A,B} & A<>B implies '/\' G = A '/\' B; theorem :: BVFUNC14:3 G={B,C,D} & B<>C & C<>D & D<>B implies '/\' G = B '/\' C '/\' D; theorem :: BVFUNC14:4 G={A,B,C} & A<>B & C<>A implies CompF(A,G) = B '/\' C; theorem :: BVFUNC14:5 G={A,B,C} & A<>B & B<>C implies CompF(B,G) = C '/\' A; theorem :: BVFUNC14:6 G={A,B,C} & B<>C & C<>A implies CompF(C,G) = A '/\' B; theorem :: BVFUNC14:7 G={A,B,C,D} & A<>B & A<>C & A<>D implies CompF(A,G) = B '/\' C '/\' D; theorem :: BVFUNC14:8 G={A,B,C,D} & A<>B & B<>C & B<>D implies CompF(B,G) = A '/\' C '/\' D; theorem :: BVFUNC14:9 G={A,B,C,D} & A<>C & B<>C & C<>D implies CompF(C,G) = A '/\' B '/\' D; theorem :: BVFUNC14:10 G={A,B,C,D} & A<>D & B<>D & C<>D implies CompF(D,G) = A '/\' C '/\' B; canceled 3; theorem :: BVFUNC14:14 for B,C,D,b,c,d being set holds dom((B .--> b) +* (C .--> c) +* (D .--> d)) = {B,C,D}; theorem :: BVFUNC14:15 for f being Function, C,D,c,d being set st C<>D holds (f +* (C .--> c) +* (D .--> d)).C = c; theorem :: BVFUNC14:16 for B,C,D,b,c,d being set st B<>C & D<>B holds ((B .--> b) +* (C .--> c) +* (D .--> d)).B = b; theorem :: BVFUNC14:17 for B,C,D,b,c,d being set, h being Function st h = (B .--> b) +* (C .--> c) +* (D .--> d) holds rng h = {h.B,h.C,h.D}; :: from BVFUNC20 theorem :: BVFUNC14:18 for h being Function, A',B',C',D' being set st G={A,B,C,D} & A<>B & A<>C & A<>D & B<>C & B<>D & C<>D & h = (B .--> B') +* (C .--> C') +* (D .--> D') +* (A .--> A') holds h.B = B' & h.C = C' & h.D = D'; theorem :: BVFUNC14:19 for A,B,C,D being set,h being Function, A',B',C',D' being set st h = (B .--> B') +* (C .--> C') +* (D .--> D') +* (A .--> A') holds dom h = {A,B,C,D}; theorem :: BVFUNC14:20 for h being Function,A',B',C',D' being set st G={A,B,C,D} & h = (B .--> B') +* (C .--> C') +* (D .--> D') +* (A .--> A') holds rng h = {h.A,h.B,h.C,h.D}; theorem :: BVFUNC14:21 for z,u being Element of Y, h being Function st G is independent & G={A,B,C,D} & A<>B & A<>C & A<>D & B<>C & B<>D & C<>D holds EqClass(u,B '/\' C '/\' D) meets EqClass(z,A); theorem :: BVFUNC14:22 for z,u being Element of Y st G is independent & G={A,B,C,D} & A<>B & A<>C & A<>D & B<>C & B<>D & C<>D & EqClass(z,C '/\' D)=EqClass(u,C '/\' D) holds EqClass(u,CompF(A,G)) meets EqClass(z,CompF(B,G)); theorem :: BVFUNC14:23 for z,u being Element of Y st G is independent & G={A,B,C} & A<>B & B<>C & C<>A & EqClass(z,C)=EqClass(u,C) holds EqClass(u,CompF(A,G)) meets EqClass(z,CompF(B,G));

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