Metamath Proof Explorer


Theorem ndmaovass

Description: Any operation is associative outside its domain. In contrast to ndmovass 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
Hypothesis ndmaov.1 dom F = S × S
Assertion ndmaovass ¬ A S B S C S A F B F C = A F B F C

Proof

Step Hyp Ref Expression
1 ndmaov.1 dom F = S × S
2 1 eleq2i A F B C dom F A F B C S × S
3 opelxp A F B C S × S A F B S C S
4 2 3 bitri A F B C dom F A F B S C S
5 aovvdm A F B S A B dom F
6 1 eleq2i A B dom F A B S × S
7 opelxp A B S × S A S B S
8 6 7 bitri A B dom F A S B S
9 df-3an A S B S C S A S B S C S
10 9 simplbi2 A S B S C S A S B S C S
11 8 10 sylbi A B dom F C S A S B S C S
12 5 11 syl A F B S C S A S B S C S
13 12 imp A F B S C S A S B S C S
14 4 13 sylbi A F B C dom F A S B S C S
15 ndmaov ¬ A F B C dom F A F B F C = V
16 14 15 nsyl5 ¬ A S B S C S A F B F C = V
17 1 eleq2i A B F C dom F A B F C S × S
18 opelxp A B F C S × S A S B F C S
19 17 18 bitri A B F C dom F A S B F C S
20 aovvdm B F C S B C dom F
21 1 eleq2i B C dom F B C S × S
22 opelxp B C S × S B S C S
23 21 22 bitri B C dom F B S C S
24 3anass A S B S C S A S B S C S
25 24 biimpri A S B S C S A S B S C S
26 25 a1d A S B S C S A B F C dom F A S B S C S
27 26 expcom B S C S A S A B F C dom F A S B S C S
28 23 27 sylbi B C dom F A S A B F C dom F A S B S C S
29 20 28 syl B F C S A S A B F C dom F A S B S C S
30 29 impcom A S B F C S A B F C dom F A S B S C S
31 19 30 sylbi A B F C dom F A B F C dom F A S B S C S
32 31 pm2.43i A B F C dom F A S B S C S
33 ndmaov ¬ A B F C dom F A F B F C = V
34 32 33 nsyl5 ¬ A S B S C S A F B F C = V
35 16 34 eqtr4d ¬ A S B S C S A F B F C = A F B F C