Metamath Proof Explorer


Theorem weisoeq

Description: Thus, there is at most one isomorphism between any two set-like well-ordered classes. Class version of wemoiso . (Contributed by Mario Carneiro, 25-Jun-2015)

Ref Expression
Assertion weisoeq R We A R Se A F Isom R , S A B G Isom R , S A B F = G

Proof

Step Hyp Ref Expression
1 id G Isom R , S A B G Isom R , S A B
2 isocnv F Isom R , S A B F -1 Isom S , R B A
3 isotr G Isom R , S A B F -1 Isom S , R B A F -1 G Isom R , R A A
4 1 2 3 syl2anr F Isom R , S A B G Isom R , S A B F -1 G Isom R , R A A
5 weniso R We A R Se A F -1 G Isom R , R A A F -1 G = I A
6 5 3expa R We A R Se A F -1 G Isom R , R A A F -1 G = I A
7 4 6 sylan2 R We A R Se A F Isom R , S A B G Isom R , S A B F -1 G = I A
8 simprl R We A R Se A F Isom R , S A B G Isom R , S A B F Isom R , S A B
9 isof1o F Isom R , S A B F : A 1-1 onto B
10 f1of1 F : A 1-1 onto B F : A 1-1 B
11 8 9 10 3syl R We A R Se A F Isom R , S A B G Isom R , S A B F : A 1-1 B
12 simprr R We A R Se A F Isom R , S A B G Isom R , S A B G Isom R , S A B
13 isof1o G Isom R , S A B G : A 1-1 onto B
14 f1of1 G : A 1-1 onto B G : A 1-1 B
15 12 13 14 3syl R We A R Se A F Isom R , S A B G Isom R , S A B G : A 1-1 B
16 f1eqcocnv F : A 1-1 B G : A 1-1 B F = G F -1 G = I A
17 11 15 16 syl2anc R We A R Se A F Isom R , S A B G Isom R , S A B F = G F -1 G = I A
18 7 17 mpbird R We A R Se A F Isom R , S A B G Isom R , S A B F = G