Metamath Proof Explorer


Theorem pprodcnveq

Description: A converse law for parallel product. (Contributed by Scott Fenton, 3-May-2014)

Ref Expression
Assertion pprodcnveq pprod R S = pprod R -1 S -1 -1

Proof

Step Hyp Ref Expression
1 dfpprod2 pprod R S = 1 st V × V -1 R 1 st V × V 2 nd V × V -1 S 2 nd V × V
2 dfpprod2 pprod R -1 S -1 = 1 st V × V -1 R -1 1 st V × V 2 nd V × V -1 S -1 2 nd V × V
3 2 cnveqi pprod R -1 S -1 -1 = 1 st V × V -1 R -1 1 st V × V 2 nd V × V -1 S -1 2 nd V × V -1
4 cnvin 1 st V × V -1 R -1 1 st V × V 2 nd V × V -1 S -1 2 nd V × V -1 = 1 st V × V -1 R -1 1 st V × V -1 2 nd V × V -1 S -1 2 nd V × V -1
5 cnvco1 1 st V × V -1 R -1 1 st V × V -1 = R -1 1 st V × V -1 1 st V × V
6 cnvco1 R -1 1 st V × V -1 = 1 st V × V -1 R
7 6 coeq1i R -1 1 st V × V -1 1 st V × V = 1 st V × V -1 R 1 st V × V
8 coass 1 st V × V -1 R 1 st V × V = 1 st V × V -1 R 1 st V × V
9 5 7 8 3eqtri 1 st V × V -1 R -1 1 st V × V -1 = 1 st V × V -1 R 1 st V × V
10 cnvco1 2 nd V × V -1 S -1 2 nd V × V -1 = S -1 2 nd V × V -1 2 nd V × V
11 cnvco1 S -1 2 nd V × V -1 = 2 nd V × V -1 S
12 11 coeq1i S -1 2 nd V × V -1 2 nd V × V = 2 nd V × V -1 S 2 nd V × V
13 coass 2 nd V × V -1 S 2 nd V × V = 2 nd V × V -1 S 2 nd V × V
14 10 12 13 3eqtri 2 nd V × V -1 S -1 2 nd V × V -1 = 2 nd V × V -1 S 2 nd V × V
15 9 14 ineq12i 1 st V × V -1 R -1 1 st V × V -1 2 nd V × V -1 S -1 2 nd V × V -1 = 1 st V × V -1 R 1 st V × V 2 nd V × V -1 S 2 nd V × V
16 3 4 15 3eqtri pprod R -1 S -1 -1 = 1 st V × V -1 R 1 st V × V 2 nd V × V -1 S 2 nd V × V
17 1 16 eqtr4i pprod R S = pprod R -1 S -1 -1