Metamath Proof Explorer


Theorem tposideq

Description: Two ways of expressing the swap function. (Contributed by Zhi Wang, 6-Oct-2025)

Ref Expression
Assertion tposideq Rel R tpos I R = x R x -1

Proof

Step Hyp Ref Expression
1 tposres Rel R tpos I R = tpos I R -1
2 relcnv Rel R -1
3 fnresi I R -1 Fn R -1
4 tposfn2 Rel R -1 I R -1 Fn R -1 tpos I R -1 Fn R -1 -1
5 2 3 4 mp2 tpos I R -1 Fn R -1 -1
6 dfrel2 Rel R R -1 -1 = R
7 6 biimpi Rel R R -1 -1 = R
8 7 fneq2d Rel R tpos I R -1 Fn R -1 -1 tpos I R -1 Fn R
9 5 8 mpbii Rel R tpos I R -1 Fn R
10 vsnex x V
11 10 cnvex x -1 V
12 11 uniex x -1 V
13 eqid x R x -1 = x R x -1
14 12 13 fnmpti x R x -1 Fn R
15 14 a1i Rel R x R x -1 Fn R
16 1st2nd Rel R y R y = 1 st y 2 nd y
17 1st2ndb y V × V y = 1 st y 2 nd y
18 17 biimpri y = 1 st y 2 nd y y V × V
19 2nd1st y V × V y -1 = 2 nd y 1 st y
20 16 18 19 3syl Rel R y R y -1 = 2 nd y 1 st y
21 sneq x = y x = y
22 21 cnveqd x = y x -1 = y -1
23 22 unieqd x = y x -1 = y -1
24 23 13 12 fvmpt3i y R x R x -1 y = y -1
25 24 adantl Rel R y R x R x -1 y = y -1
26 16 fveq2d Rel R y R tpos I R -1 y = tpos I R -1 1 st y 2 nd y
27 ovtpos 1 st y tpos I R -1 2 nd y = 2 nd y I R -1 1 st y
28 df-ov 1 st y tpos I R -1 2 nd y = tpos I R -1 1 st y 2 nd y
29 df-ov 2 nd y I R -1 1 st y = I R -1 2 nd y 1 st y
30 27 28 29 3eqtr3i tpos I R -1 1 st y 2 nd y = I R -1 2 nd y 1 st y
31 30 a1i Rel R y R tpos I R -1 1 st y 2 nd y = I R -1 2 nd y 1 st y
32 simpr Rel R y R y R
33 16 32 eqeltrrd Rel R y R 1 st y 2 nd y R
34 fvex 2 nd y V
35 fvex 1 st y V
36 34 35 opelcnv 2 nd y 1 st y R -1 1 st y 2 nd y R
37 36 biimpri 1 st y 2 nd y R 2 nd y 1 st y R -1
38 fvresi 2 nd y 1 st y R -1 I R -1 2 nd y 1 st y = 2 nd y 1 st y
39 33 37 38 3syl Rel R y R I R -1 2 nd y 1 st y = 2 nd y 1 st y
40 26 31 39 3eqtrd Rel R y R tpos I R -1 y = 2 nd y 1 st y
41 20 25 40 3eqtr4rd Rel R y R tpos I R -1 y = x R x -1 y
42 9 15 41 eqfnfvd Rel R tpos I R -1 = x R x -1
43 1 42 eqtrd Rel R tpos I R = x R x -1