Metamath Proof Explorer


Theorem f12dfv

Description: A one-to-one function with a domain with at least two different elements in terms of function values. (Contributed by Alexander van der Vekens, 2-Mar-2018)

Ref Expression
Hypothesis f12dfv.a A=XY
Assertion f12dfv XUYVXYF:A1-1BF:ABFXFY

Proof

Step Hyp Ref Expression
1 f12dfv.a A=XY
2 dff14b F:A1-1BF:ABxAyAxFxFy
3 1 raleqi xAyAxFxFyxXYyAxFxFy
4 sneq x=Xx=X
5 4 difeq2d x=XAx=AX
6 fveq2 x=XFx=FX
7 6 neeq1d x=XFxFyFXFy
8 5 7 raleqbidv x=XyAxFxFyyAXFXFy
9 sneq x=Yx=Y
10 9 difeq2d x=YAx=AY
11 fveq2 x=YFx=FY
12 11 neeq1d x=YFxFyFYFy
13 10 12 raleqbidv x=YyAxFxFyyAYFYFy
14 8 13 ralprg XUYVxXYyAxFxFyyAXFXFyyAYFYFy
15 14 adantr XUYVXYxXYyAxFxFyyAXFXFyyAYFYFy
16 1 difeq1i AX=XYX
17 difprsn1 XYXYX=Y
18 16 17 eqtrid XYAX=Y
19 18 adantl XUYVXYAX=Y
20 19 raleqdv XUYVXYyAXFXFyyYFXFy
21 fveq2 y=YFy=FY
22 21 neeq2d y=YFXFyFXFY
23 22 ralsng YVyYFXFyFXFY
24 23 adantl XUYVyYFXFyFXFY
25 24 adantr XUYVXYyYFXFyFXFY
26 20 25 bitrd XUYVXYyAXFXFyFXFY
27 1 difeq1i AY=XYY
28 difprsn2 XYXYY=X
29 27 28 eqtrid XYAY=X
30 29 adantl XUYVXYAY=X
31 30 raleqdv XUYVXYyAYFYFyyXFYFy
32 fveq2 y=XFy=FX
33 32 neeq2d y=XFYFyFYFX
34 33 ralsng XUyXFYFyFYFX
35 34 adantr XUYVyXFYFyFYFX
36 35 adantr XUYVXYyXFYFyFYFX
37 31 36 bitrd XUYVXYyAYFYFyFYFX
38 26 37 anbi12d XUYVXYyAXFXFyyAYFYFyFXFYFYFX
39 necom FXFYFYFX
40 39 biimpi FXFYFYFX
41 40 pm4.71i FXFYFXFYFYFX
42 38 41 bitr4di XUYVXYyAXFXFyyAYFYFyFXFY
43 15 42 bitrd XUYVXYxXYyAxFxFyFXFY
44 3 43 bitrid XUYVXYxAyAxFxFyFXFY
45 44 anbi2d XUYVXYF:ABxAyAxFxFyF:ABFXFY
46 2 45 bitrid XUYVXYF:A1-1BF:ABFXFY