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

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

let a, b be Function of Y,BOOLEAN; :: thesis: for PA being a_partition of Y holds All (a,PA,G) '<' (Ex (b,PA,G)) 'imp' (Ex ((a '&' b),PA,G))
let PA be a_partition of Y; :: thesis: All (a,PA,G) '<' (Ex (b,PA,G)) 'imp' (Ex ((a '&' b),PA,G))
let z be Element of Y; :: according to BVFUNC_1:def 12 :: thesis: ( not (All (a,PA,G)) . z = TRUE or ((Ex (b,PA,G)) 'imp' (Ex ((a '&' b),PA,G))) . z = TRUE )
assume A1: (All (a,PA,G)) . z = TRUE ; :: thesis: ((Ex (b,PA,G)) 'imp' (Ex ((a '&' b),PA,G))) . z = TRUE
A2: now :: thesis: for x being Element of Y st x in EqClass (z,(CompF (PA,G))) holds
a . x = TRUE
assume ex x being Element of Y st
( x in EqClass (z,(CompF (PA,G))) & not a . x = TRUE ) ; :: thesis: contradiction
then (B_INF (a,(CompF (PA,G)))) . z = FALSE by BVFUNC_1:def 16;
hence contradiction by A1, BVFUNC_2:def 9; :: thesis: verum
end;
per cases ( (Ex (b,PA,G)) . z = TRUE or (Ex (b,PA,G)) . z <> TRUE ) ;
suppose A3: (Ex (b,PA,G)) . z = TRUE ; :: thesis: ((Ex (b,PA,G)) 'imp' (Ex ((a '&' b),PA,G))) . z = TRUE
now :: thesis: ex x being Element of Y st
( x in EqClass (z,(CompF (PA,G))) & b . x = TRUE )
assume for x being Element of Y holds
( not x in EqClass (z,(CompF (PA,G))) or not b . x = TRUE ) ; :: thesis: contradiction
then (B_SUP (b,(CompF (PA,G)))) . z = FALSE by BVFUNC_1:def 17;
hence contradiction by A3, BVFUNC_2:def 10; :: thesis: verum
end;
then consider x1 being Element of Y such that
A4: x1 in EqClass (z,(CompF (PA,G))) and
A5: b . x1 = TRUE ;
(a '&' b) . x1 = (a . x1) '&' (b . x1) by MARGREL1:def 20
.= TRUE '&' TRUE by A2, A4, A5
.= TRUE ;
then (B_SUP ((a '&' b),(CompF (PA,G)))) . z = TRUE by ;
then (Ex ((a '&' b),PA,G)) . z = TRUE by BVFUNC_2:def 10;
hence ((Ex (b,PA,G)) 'imp' (Ex ((a '&' b),PA,G))) . z = ('not' ((Ex (b,PA,G)) . z)) 'or' TRUE by BVFUNC_1:def 8
.= TRUE by BINARITH:10 ;
:: thesis: verum
end;
suppose (Ex (b,PA,G)) . z <> TRUE ; :: thesis: ((Ex (b,PA,G)) 'imp' (Ex ((a '&' b),PA,G))) . z = TRUE
then (Ex (b,PA,G)) . z = FALSE by XBOOLEAN:def 3;
hence ((Ex (b,PA,G)) 'imp' (Ex ((a '&' b),PA,G))) . z = () 'or' ((Ex ((a '&' b),PA,G)) . z) by BVFUNC_1:def 8
.= TRUE 'or' ((Ex ((a '&' b),PA,G)) . z) by MARGREL1:11
.= TRUE by BINARITH:10 ;
:: thesis: verum
end;
end;