let Y be non empty set ; :: thesis: for G being Subset of ()
for a, b, c being Function of Y,BOOLEAN
for PA being a_partition of Y holds (All ((a 'imp' c),PA,G)) '&' (All ((c 'imp' b),PA,G)) '<' All ((a 'imp' b),PA,G)

let G be Subset of (); :: thesis: for a, b, c being Function of Y,BOOLEAN
for PA being a_partition of Y holds (All ((a 'imp' c),PA,G)) '&' (All ((c 'imp' b),PA,G)) '<' All ((a 'imp' b),PA,G)

let a, b, c be Function of Y,BOOLEAN; :: thesis: for PA being a_partition of Y holds (All ((a 'imp' c),PA,G)) '&' (All ((c 'imp' b),PA,G)) '<' All ((a 'imp' b),PA,G)
let PA be a_partition of Y; :: thesis: (All ((a 'imp' c),PA,G)) '&' (All ((c 'imp' b),PA,G)) '<' All ((a 'imp' b),PA,G)
let z be Element of Y; :: according to BVFUNC_1:def 12 :: thesis: ( not ((All ((a 'imp' c),PA,G)) '&' (All ((c 'imp' b),PA,G))) . z = TRUE or (All ((a 'imp' b),PA,G)) . z = TRUE )
assume ((All ((a 'imp' c),PA,G)) '&' (All ((c 'imp' b),PA,G))) . z = TRUE ; :: thesis: (All ((a 'imp' b),PA,G)) . z = TRUE
then A1: ((All ((a 'imp' c),PA,G)) . z) '&' ((All ((c 'imp' b),PA,G)) . z) = TRUE by MARGREL1:def 20;
A2: now :: thesis: for x being Element of Y st x in EqClass (z,(CompF (PA,G))) holds
(c 'imp' b) . x = TRUE
assume ex x being Element of Y st
( x in EqClass (z,(CompF (PA,G))) & not (c 'imp' b) . x = TRUE ) ; :: thesis: contradiction
then (B_INF ((c 'imp' b),(CompF (PA,G)))) . z = FALSE by BVFUNC_1:def 16;
then (All ((c 'imp' b),PA,G)) . z = FALSE by BVFUNC_2:def 9;
hence contradiction by A1, MARGREL1:12; :: thesis: verum
end;
A3: now :: thesis: for x being Element of Y st x in EqClass (z,(CompF (PA,G))) holds
(a 'imp' c) . x = TRUE
assume ex x being Element of Y st
( x in EqClass (z,(CompF (PA,G))) & not (a 'imp' c) . x = TRUE ) ; :: thesis: contradiction
then (B_INF ((a 'imp' c),(CompF (PA,G)))) . z = FALSE by BVFUNC_1:def 16;
then (All ((a 'imp' c),PA,G)) . z = FALSE by BVFUNC_2:def 9;
hence contradiction by A1, MARGREL1:12; :: thesis: verum
end;
for x being Element of Y st x in EqClass (z,(CompF (PA,G))) holds
(a 'imp' b) . x = TRUE
proof
let x be Element of Y; :: thesis: ( x in EqClass (z,(CompF (PA,G))) implies (a 'imp' b) . x = TRUE )
A4: ( 'not' (a . x) = TRUE or 'not' (a . x) = FALSE ) by XBOOLEAN:def 3;
A5: ( 'not' (c . x) = TRUE or 'not' (c . x) = FALSE ) by XBOOLEAN:def 3;
assume A6: x in EqClass (z,(CompF (PA,G))) ; :: thesis: (a 'imp' b) . x = TRUE
then (a 'imp' c) . x = TRUE by A3;
then A7: ('not' (a . x)) 'or' (c . x) = TRUE by BVFUNC_1:def 8;
(c 'imp' b) . x = TRUE by A2, A6;
then A8: ('not' (c . x)) 'or' (b . x) = TRUE by BVFUNC_1:def 8;
per cases ( ( 'not' (a . x) = TRUE & 'not' (c . x) = TRUE ) or ( 'not' (a . x) = TRUE & b . x = TRUE ) or ( c . x = TRUE & 'not' (c . x) = TRUE ) or ( c . x = TRUE & b . x = TRUE ) ) by ;
suppose ( 'not' (a . x) = TRUE & 'not' (c . x) = TRUE ) ; :: thesis: (a 'imp' b) . x = TRUE
hence (a 'imp' b) . x = TRUE 'or' (b . x) by BVFUNC_1:def 8
.= TRUE by BINARITH:10 ;
:: thesis: verum
end;
suppose ( 'not' (a . x) = TRUE & b . x = TRUE ) ; :: thesis: (a 'imp' b) . x = TRUE
hence (a 'imp' b) . x = TRUE 'or' (b . x) by BVFUNC_1:def 8
.= TRUE by BINARITH:10 ;
:: thesis: verum
end;
suppose ( c . x = TRUE & 'not' (c . x) = TRUE ) ; :: thesis: (a 'imp' b) . x = TRUE
end;
suppose ( c . x = TRUE & b . x = TRUE ) ; :: thesis: (a 'imp' b) . x = TRUE
hence (a 'imp' b) . x = ('not' (a . x)) 'or' TRUE by BVFUNC_1:def 8
.= TRUE by BINARITH:10 ;
:: thesis: verum
end;
end;
end;
then (B_INF ((a 'imp' b),(CompF (PA,G)))) . z = TRUE by BVFUNC_1:def 16;
hence (All ((a 'imp' b),PA,G)) . z = TRUE by BVFUNC_2:def 9; :: thesis: verum