Metamath Proof Explorer

Theorem ndmaovdistr

Description: Any operation is distributive outside its domain. In contrast to ndmovdistr where it is required that the operation's domain doesn't contain the empty set ( -. (/) e. S ), no additional assumption is required. (Contributed by Alexander van der Vekens, 26-May-2017)

Ref Expression
Hypotheses ndmaov.1 
|- dom F = ( S X. S )
ndmaov.6 
|- dom G = ( S X. S )
Assertion ndmaovdistr 
|- ( -. ( A e. S /\ B e. S /\ C e. S ) -> (( A G (( B F C )) )) = (( (( A G B )) F (( A G C )) )) )

Proof

Step Hyp Ref Expression
1 ndmaov.1 
 |-  dom F = ( S X. S )
2 ndmaov.6 
 |-  dom G = ( S X. S )
3 2 eleq2i 
 |-  ( <. A , (( B F C )) >. e. dom G <-> <. A , (( B F C )) >. e. ( S X. S ) )
4 opelxp 
 |-  ( <. A , (( B F C )) >. e. ( S X. S ) <-> ( A e. S /\ (( B F C )) e. S ) )
5 3 4 bitri 
 |-  ( <. A , (( B F C )) >. e. dom G <-> ( A e. S /\ (( B F C )) e. S ) )
6 aovvdm 
 |-  ( (( B F C )) e. S -> <. B , C >. e. dom F )
7 1 eleq2i 
 |-  ( <. B , C >. e. dom F <-> <. B , C >. e. ( S X. S ) )
8 opelxp 
 |-  ( <. B , C >. e. ( S X. S ) <-> ( B e. S /\ C e. S ) )
9 7 8 bitri 
 |-  ( <. B , C >. e. dom F <-> ( B e. S /\ C e. S ) )
10 3anass 
 |-  ( ( A e. S /\ B e. S /\ C e. S ) <-> ( A e. S /\ ( B e. S /\ C e. S ) ) )
11 10 simplbi2com 
 |-  ( ( B e. S /\ C e. S ) -> ( A e. S -> ( A e. S /\ B e. S /\ C e. S ) ) )
12 9 11 sylbi 
 |-  ( <. B , C >. e. dom F -> ( A e. S -> ( A e. S /\ B e. S /\ C e. S ) ) )
13 6 12 syl 
 |-  ( (( B F C )) e. S -> ( A e. S -> ( A e. S /\ B e. S /\ C e. S ) ) )
14 13 impcom 
 |-  ( ( A e. S /\ (( B F C )) e. S ) -> ( A e. S /\ B e. S /\ C e. S ) )
15 5 14 sylbi 
 |-  ( <. A , (( B F C )) >. e. dom G -> ( A e. S /\ B e. S /\ C e. S ) )
16 ndmaov 
 |-  ( -. <. A , (( B F C )) >. e. dom G -> (( A G (( B F C )) )) = _V )
17 15 16 nsyl5 
 |-  ( -. ( A e. S /\ B e. S /\ C e. S ) -> (( A G (( B F C )) )) = _V )
18 1 eleq2i 
 |-  ( <. (( A G B )) , (( A G C )) >. e. dom F <-> <. (( A G B )) , (( A G C )) >. e. ( S X. S ) )
19 opelxp 
 |-  ( <. (( A G B )) , (( A G C )) >. e. ( S X. S ) <-> ( (( A G B )) e. S /\ (( A G C )) e. S ) )
20 18 19 bitri 
 |-  ( <. (( A G B )) , (( A G C )) >. e. dom F <-> ( (( A G B )) e. S /\ (( A G C )) e. S ) )
21 aovvdm 
 |-  ( (( A G B )) e. S -> <. A , B >. e. dom G )
22 2 eleq2i 
 |-  ( <. A , B >. e. dom G <-> <. A , B >. e. ( S X. S ) )
23 opelxp 
 |-  ( <. A , B >. e. ( S X. S ) <-> ( A e. S /\ B e. S ) )
24 22 23 bitri 
 |-  ( <. A , B >. e. dom G <-> ( A e. S /\ B e. S ) )
25 2 eleq2i 
 |-  ( <. A , C >. e. dom G <-> <. A , C >. e. ( S X. S ) )
26 opelxp 
 |-  ( <. A , C >. e. ( S X. S ) <-> ( A e. S /\ C e. S ) )
27 25 26 bitri 
 |-  ( <. A , C >. e. dom G <-> ( A e. S /\ C e. S ) )
28 simpll 
 |-  ( ( ( A e. S /\ C e. S ) /\ ( A e. S /\ B e. S ) ) -> A e. S )
29 simprr 
 |-  ( ( ( A e. S /\ C e. S ) /\ ( A e. S /\ B e. S ) ) -> B e. S )
30 simplr 
 |-  ( ( ( A e. S /\ C e. S ) /\ ( A e. S /\ B e. S ) ) -> C e. S )
31 28 29 30 3jca 
 |-  ( ( ( A e. S /\ C e. S ) /\ ( A e. S /\ B e. S ) ) -> ( A e. S /\ B e. S /\ C e. S ) )
32 31 ex 
 |-  ( ( A e. S /\ C e. S ) -> ( ( A e. S /\ B e. S ) -> ( A e. S /\ B e. S /\ C e. S ) ) )
33 27 32 sylbi 
 |-  ( <. A , C >. e. dom G -> ( ( A e. S /\ B e. S ) -> ( A e. S /\ B e. S /\ C e. S ) ) )
34 aovvdm 
 |-  ( (( A G C )) e. S -> <. A , C >. e. dom G )
35 33 34 syl11 
 |-  ( ( A e. S /\ B e. S ) -> ( (( A G C )) e. S -> ( A e. S /\ B e. S /\ C e. S ) ) )
36 24 35 sylbi 
 |-  ( <. A , B >. e. dom G -> ( (( A G C )) e. S -> ( A e. S /\ B e. S /\ C e. S ) ) )
37 21 36 syl 
 |-  ( (( A G B )) e. S -> ( (( A G C )) e. S -> ( A e. S /\ B e. S /\ C e. S ) ) )
38 37 imp 
 |-  ( ( (( A G B )) e. S /\ (( A G C )) e. S ) -> ( A e. S /\ B e. S /\ C e. S ) )
39 20 38 sylbi 
 |-  ( <. (( A G B )) , (( A G C )) >. e. dom F -> ( A e. S /\ B e. S /\ C e. S ) )
40 ndmaov 
 |-  ( -. <. (( A G B )) , (( A G C )) >. e. dom F -> (( (( A G B )) F (( A G C )) )) = _V )
41 39 40 nsyl5 
 |-  ( -. ( A e. S /\ B e. S /\ C e. S ) -> (( (( A G B )) F (( A G C )) )) = _V )
42 17 41 eqtr4d 
 |-  ( -. ( A e. S /\ B e. S /\ C e. S ) -> (( A G (( B F C )) )) = (( (( A G B )) F (( A G C )) )) )