let Y be non empty set ; :: thesis: for G being Subset of (PARTITIONS Y)
for a, b being Element of Funcs Y,BOOLEAN
for PA being a_partition of Y holds All (a 'imp' b),PA,G = All (('not' a) 'or' b),PA,G
let G be Subset of (PARTITIONS Y); :: thesis: for a, b being Element of Funcs Y,BOOLEAN
for PA being a_partition of Y holds All (a 'imp' b),PA,G = All (('not' a) 'or' b),PA,G
let a, b be Element of Funcs Y,BOOLEAN ; :: thesis: for PA being a_partition of Y holds All (a 'imp' b),PA,G = All (('not' a) 'or' b),PA,G
let PA be a_partition of Y; :: thesis: All (a 'imp' b),PA,G = All (('not' a) 'or' b),PA,G
A1: All (('not' a) 'or' b),PA,G '<' All (a 'imp' b),PA,G

proof

let z be Element of Y; :: according to BVFUNC_1:def 15 :: thesis: ( not (All (('not' a) 'or' b),PA,G) . z = TRUE or (All (a 'imp' b),PA,G) . z = TRUE )
assume A2: (All (('not' a) 'or' b),PA,G) . z = TRUE ; :: thesis: (All (a 'imp' b),PA,G) . z = TRUE

A3: now

assume ex x being Element of Y st
( x in EqClass z,(CompF PA,G) & not (('not' a) 'or' b) . x = TRUE ) ; :: thesis: contradiction
then (B_INF (('not' a) 'or' b),(CompF PA,G)) . z = FALSE by BVFUNC_1:def 19;
hence contradiction by A2, BVFUNC_2:def 9; :: 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;
assume x in EqClass z,(CompF PA,G) ; :: thesis: (a 'imp' b) . x = TRUE
then (('not' a) 'or' b) . x = TRUE by A3;
then A5: (('not' a) . x) 'or' (b . x) = TRUE by BVFUNC_1:def 7;

per cases ( ('not' a) . x = TRUE or b . x = TRUE ) by A5, A4, BINARITH:7;

suppose ('not' a) . x = TRUE ; :: thesis: (a 'imp' b) . x = TRUE

then 'not' (a . x) = TRUE by MARGREL1:def 20;
hence (a 'imp' b) . x = TRUE 'or' (b . x) by BVFUNC_1:def 11
.= TRUE by BINARITH:19 ;
:: thesis: verum

end;

suppose b . x = TRUE ; :: thesis: (a 'imp' b) . x = TRUE

hence (a 'imp' b) . x = ('not' (a . x)) 'or' TRUE by BVFUNC_1:def 11
.= TRUE by BINARITH:19 ;
:: thesis: verum

end;

end;

end;

then (B_INF (a 'imp' b),(CompF PA,G)) . z = TRUE by BVFUNC_1:def 19;
hence (All (a 'imp' b),PA,G) . z = TRUE by BVFUNC_2:def 9; :: thesis: verum

end;

All (a 'imp' b),PA,G '<' All (('not' a) 'or' b),PA,G

proof

let z be Element of Y; :: according to BVFUNC_1:def 15 :: thesis: ( not (All (a 'imp' b),PA,G) . z = TRUE or (All (('not' a) 'or' b),PA,G) . z = TRUE )
assume A6: (All (a 'imp' b),PA,G) . z = TRUE ; :: thesis: (All (('not' a) 'or' b),PA,G) . z = TRUE

A7: now

assume ex x being Element of Y st
( x in EqClass z,(CompF PA,G) & not (a 'imp' b) . x = TRUE ) ; :: thesis: contradiction
then (B_INF (a 'imp' b),(CompF PA,G)) . z = FALSE by BVFUNC_1:def 19;
hence contradiction by A6, BVFUNC_2:def 9; :: thesis: verum

end;

now

let x be Element of Y; :: thesis: ( x in EqClass z,(CompF PA,G) implies (('not' a) 'or' b) . b₁ = TRUE )
A8: ( 'not' (a . x) = TRUE or 'not' (a . x) = FALSE ) by XBOOLEAN:def 3;
assume x in EqClass z,(CompF PA,G) ; :: thesis: (('not' a) 'or' b) . b₁ = TRUE
then (a 'imp' b) . x = TRUE by A7;
then A9: ('not' (a . x)) 'or' (b . x) = TRUE by BVFUNC_1:def 11;

per cases ( 'not' (a . x) = TRUE or b . x = TRUE ) by A9, A8, BINARITH:7;

suppose 'not' (a . x) = TRUE ; :: thesis: (('not' a) 'or' b) . b₁ = TRUE

then ('not' a) . x = TRUE by MARGREL1:def 20;
hence (('not' a) 'or' b) . x = TRUE 'or' (b . x) by BVFUNC_1:def 7
.= TRUE by BINARITH:19 ;
:: thesis: verum

end;

suppose b . x = TRUE ; :: thesis: (('not' a) 'or' b) . b₁ = TRUE

hence (('not' a) 'or' b) . x = (('not' a) . x) 'or' TRUE by BVFUNC_1:def 7
.= TRUE by BINARITH:19 ;
:: thesis: verum

end;

end;

end;

then (B_INF (('not' a) 'or' b),(CompF PA,G)) . z = TRUE by BVFUNC_1:def 19;
hence (All (('not' a) 'or' b),PA,G) . z = TRUE by BVFUNC_2:def 9; :: thesis: verum

end;

hence All (a 'imp' b),PA,G = All (('not' a) 'or' b),PA,G by A1, BVFUNC_1:18; :: thesis: verum