Metamath Proof Explorer


Theorem 2f1fvneq

Description: If two one-to-one functions are applied on different arguments, also the values are different. (Contributed by Alexander van der Vekens, 25-Jan-2018) (Proof shortened by AV, 30-Oct-2025)

Ref Expression
Assertion 2f1fvneq E : D 1-1 R F : C 1-1 D A C B C A B E F A = X E F B = Y X Y

Proof

Step Hyp Ref Expression
1 simp1l E : D 1-1 R F : C 1-1 D A C B C A B E : D 1-1 R
2 f1f F : C 1-1 D F : C D
3 2 adantl E : D 1-1 R F : C 1-1 D F : C D
4 3 adantr E : D 1-1 R F : C 1-1 D A C B C F : C D
5 simpl A C B C A C
6 5 adantl E : D 1-1 R F : C 1-1 D A C B C A C
7 4 6 ffvelcdmd E : D 1-1 R F : C 1-1 D A C B C F A D
8 7 3adant3 E : D 1-1 R F : C 1-1 D A C B C A B F A D
9 simpr A C B C B C
10 9 adantl E : D 1-1 R F : C 1-1 D A C B C B C
11 4 10 ffvelcdmd E : D 1-1 R F : C 1-1 D A C B C F B D
12 11 3adant3 E : D 1-1 R F : C 1-1 D A C B C A B F B D
13 simpr E : D 1-1 R F : C 1-1 D F : C 1-1 D
14 df-3an A C B C A B A C B C A B
15 14 biimpri A C B C A B A C B C A B
16 dff14i F : C 1-1 D A C B C A B F A F B
17 13 15 16 syl3an132 E : D 1-1 R F : C 1-1 D A C B C A B F A F B
18 dff14i E : D 1-1 R F A D F B D F A F B E F A E F B
19 1 8 12 17 18 syl13anc E : D 1-1 R F : C 1-1 D A C B C A B E F A E F B
20 simpl E F A = X E F B = Y E F A = X
21 simpr E F A = X E F B = Y E F B = Y
22 20 21 neeq12d E F A = X E F B = Y E F A E F B X Y
23 19 22 syl5ibcom E : D 1-1 R F : C 1-1 D A C B C A B E F A = X E F B = Y X Y