set x = q `1 ;
set y = q `2 ;
set a = q `3 ;
let f1, f2 be BiFunction of (new_set2 A),L; ( ( for u, v being Element of A holds f1 . u,v = d . u,v ) & f1 . {A},{A} = Bottom L & f1 . {{A}},{{A}} = Bottom L & f1 . {A},{{A}} = ((d . (q `1 ),(q `2 )) "\/" (q `3 )) "/\" (q `4 ) & f1 . {{A}},{A} = ((d . (q `1 ),(q `2 )) "\/" (q `3 )) "/\" (q `4 ) & ( for u being Element of A holds
( f1 . u,{A} = (d . u,(q `1 )) "\/" (q `3 ) & f1 . {A},u = (d . u,(q `1 )) "\/" (q `3 ) & f1 . u,{{A}} = (d . u,(q `2 )) "\/" (q `3 ) & f1 . {{A}},u = (d . u,(q `2 )) "\/" (q `3 ) ) ) & ( for u, v being Element of A holds f2 . u,v = d . u,v ) & f2 . {A},{A} = Bottom L & f2 . {{A}},{{A}} = Bottom L & f2 . {A},{{A}} = ((d . (q `1 ),(q `2 )) "\/" (q `3 )) "/\" (q `4 ) & f2 . {{A}},{A} = ((d . (q `1 ),(q `2 )) "\/" (q `3 )) "/\" (q `4 ) & ( for u being Element of A holds
( f2 . u,{A} = (d . u,(q `1 )) "\/" (q `3 ) & f2 . {A},u = (d . u,(q `1 )) "\/" (q `3 ) & f2 . u,{{A}} = (d . u,(q `2 )) "\/" (q `3 ) & f2 . {{A}},u = (d . u,(q `2 )) "\/" (q `3 ) ) ) implies f1 = f2 )
assume that
A20:
for u, v being Element of A holds f1 . u,v = d . u,v
and
A21:
f1 . {A},{A} = Bottom L
and
A22:
f1 . {{A}},{{A}} = Bottom L
and
A23:
f1 . {A},{{A}} = ((d . (q `1 ),(q `2 )) "\/" (q `3 )) "/\" (q `4 )
and
A24:
f1 . {{A}},{A} = ((d . (q `1 ),(q `2 )) "\/" (q `3 )) "/\" (q `4 )
and
A25:
for u being Element of A holds
( f1 . u,{A} = (d . u,(q `1 )) "\/" (q `3 ) & f1 . {A},u = (d . u,(q `1 )) "\/" (q `3 ) & f1 . u,{{A}} = (d . u,(q `2 )) "\/" (q `3 ) & f1 . {{A}},u = (d . u,(q `2 )) "\/" (q `3 ) )
and
A26:
for u, v being Element of A holds f2 . u,v = d . u,v
and
A27:
f2 . {A},{A} = Bottom L
and
A28:
f2 . {{A}},{{A}} = Bottom L
and
A29:
f2 . {A},{{A}} = ((d . (q `1 ),(q `2 )) "\/" (q `3 )) "/\" (q `4 )
and
A30:
f2 . {{A}},{A} = ((d . (q `1 ),(q `2 )) "\/" (q `3 )) "/\" (q `4 )
and
A31:
for u being Element of A holds
( f2 . u,{A} = (d . u,(q `1 )) "\/" (q `3 ) & f2 . {A},u = (d . u,(q `1 )) "\/" (q `3 ) & f2 . u,{{A}} = (d . u,(q `2 )) "\/" (q `3 ) & f2 . {{A}},u = (d . u,(q `2 )) "\/" (q `3 ) )
; f1 = f2
hence
f1 = f2
by BINOP_1:2; verum