Metamath Proof Explorer


Theorem elghomlem1OLD

Description: Obsolete as of 15-Mar-2020. Lemma for elghomOLD . (Contributed by Paul Chapman, 25-Feb-2008) (New usage is discouraged.) (Proof modification is discouraged.)

Ref Expression
Hypothesis elghomlem1OLD.1 S = f | f : ran G ran H x ran G y ran G f x H f y = f x G y
Assertion elghomlem1OLD G GrpOp H GrpOp G GrpOpHom H = S

Proof

Step Hyp Ref Expression
1 elghomlem1OLD.1 S = f | f : ran G ran H x ran G y ran G f x H f y = f x G y
2 rnexg G GrpOp ran G V
3 rnexg H GrpOp ran H V
4 1 fabexg ran G V ran H V S V
5 2 3 4 syl2an G GrpOp H GrpOp S V
6 rneq g = G ran g = ran G
7 6 feq2d g = G f : ran g ran h f : ran G ran h
8 oveq g = G x g y = x G y
9 8 fveq2d g = G f x g y = f x G y
10 9 eqeq2d g = G f x h f y = f x g y f x h f y = f x G y
11 6 10 raleqbidv g = G y ran g f x h f y = f x g y y ran G f x h f y = f x G y
12 6 11 raleqbidv g = G x ran g y ran g f x h f y = f x g y x ran G y ran G f x h f y = f x G y
13 7 12 anbi12d g = G f : ran g ran h x ran g y ran g f x h f y = f x g y f : ran G ran h x ran G y ran G f x h f y = f x G y
14 13 abbidv g = G f | f : ran g ran h x ran g y ran g f x h f y = f x g y = f | f : ran G ran h x ran G y ran G f x h f y = f x G y
15 rneq h = H ran h = ran H
16 15 feq3d h = H f : ran G ran h f : ran G ran H
17 oveq h = H f x h f y = f x H f y
18 17 eqeq1d h = H f x h f y = f x G y f x H f y = f x G y
19 18 2ralbidv h = H x ran G y ran G f x h f y = f x G y x ran G y ran G f x H f y = f x G y
20 16 19 anbi12d h = H f : ran G ran h x ran G y ran G f x h f y = f x G y f : ran G ran H x ran G y ran G f x H f y = f x G y
21 20 abbidv h = H f | f : ran G ran h x ran G y ran G f x h f y = f x G y = f | f : ran G ran H x ran G y ran G f x H f y = f x G y
22 21 1 eqtr4di h = H f | f : ran G ran h x ran G y ran G f x h f y = f x G y = S
23 df-ghomOLD GrpOpHom = g GrpOp , h GrpOp f | f : ran g ran h x ran g y ran g f x h f y = f x g y
24 14 22 23 ovmpog G GrpOp H GrpOp S V G GrpOpHom H = S
25 5 24 mpd3an3 G GrpOp H GrpOp G GrpOpHom H = S