Metamath Proof Explorer


Theorem cnvf1olem

Description: Lemma for cnvf1o . (Contributed by Mario Carneiro, 27-Apr-2014)

Ref Expression
Assertion cnvf1olem Rel A B A C = B -1 C A -1 B = C -1

Proof

Step Hyp Ref Expression
1 simprr Rel A B A C = B -1 C = B -1
2 1st2nd Rel A B A B = 1 st B 2 nd B
3 2 adantrr Rel A B A C = B -1 B = 1 st B 2 nd B
4 3 sneqd Rel A B A C = B -1 B = 1 st B 2 nd B
5 4 cnveqd Rel A B A C = B -1 B -1 = 1 st B 2 nd B -1
6 5 unieqd Rel A B A C = B -1 B -1 = 1 st B 2 nd B -1
7 1 6 eqtrd Rel A B A C = B -1 C = 1 st B 2 nd B -1
8 opswap 1 st B 2 nd B -1 = 2 nd B 1 st B
9 7 8 eqtrdi Rel A B A C = B -1 C = 2 nd B 1 st B
10 simprl Rel A B A C = B -1 B A
11 3 10 eqeltrrd Rel A B A C = B -1 1 st B 2 nd B A
12 fvex 2 nd B V
13 fvex 1 st B V
14 12 13 opelcnv 2 nd B 1 st B A -1 1 st B 2 nd B A
15 11 14 sylibr Rel A B A C = B -1 2 nd B 1 st B A -1
16 9 15 eqeltrd Rel A B A C = B -1 C A -1
17 opswap 2 nd B 1 st B -1 = 1 st B 2 nd B
18 17 eqcomi 1 st B 2 nd B = 2 nd B 1 st B -1
19 9 sneqd Rel A B A C = B -1 C = 2 nd B 1 st B
20 19 cnveqd Rel A B A C = B -1 C -1 = 2 nd B 1 st B -1
21 20 unieqd Rel A B A C = B -1 C -1 = 2 nd B 1 st B -1
22 18 3 21 3eqtr4a Rel A B A C = B -1 B = C -1
23 16 22 jca Rel A B A C = B -1 C A -1 B = C -1