Metamath Proof Explorer


Theorem grpoidval

Description: Lemma for grpoidcl and others. (Contributed by NM, 5-Feb-2010) (Proof shortened by Mario Carneiro, 15-Dec-2013) (New usage is discouraged.)

Ref Expression
Hypotheses grpoidval.1 X = ran G
grpoidval.2 U = GId G
Assertion grpoidval G GrpOp U = ι u X | x X u G x = x

Proof

Step Hyp Ref Expression
1 grpoidval.1 X = ran G
2 grpoidval.2 U = GId G
3 1 gidval G GrpOp GId G = ι u X | x X u G x = x x G u = x
4 simpl u G x = x x G u = x u G x = x
5 4 ralimi x X u G x = x x G u = x x X u G x = x
6 5 rgenw u X x X u G x = x x G u = x x X u G x = x
7 6 a1i G GrpOp u X x X u G x = x x G u = x x X u G x = x
8 1 grpoidinv G GrpOp u X x X u G x = x x G u = x y X y G x = u x G y = u
9 simpl u G x = x x G u = x y X y G x = u x G y = u u G x = x x G u = x
10 9 ralimi x X u G x = x x G u = x y X y G x = u x G y = u x X u G x = x x G u = x
11 10 reximi u X x X u G x = x x G u = x y X y G x = u x G y = u u X x X u G x = x x G u = x
12 8 11 syl G GrpOp u X x X u G x = x x G u = x
13 1 grpoideu G GrpOp ∃! u X x X u G x = x
14 7 12 13 3jca G GrpOp u X x X u G x = x x G u = x x X u G x = x u X x X u G x = x x G u = x ∃! u X x X u G x = x
15 reupick2 u X x X u G x = x x G u = x x X u G x = x u X x X u G x = x x G u = x ∃! u X x X u G x = x u X x X u G x = x x X u G x = x x G u = x
16 14 15 sylan G GrpOp u X x X u G x = x x X u G x = x x G u = x
17 16 riotabidva G GrpOp ι u X | x X u G x = x = ι u X | x X u G x = x x G u = x
18 3 17 eqtr4d G GrpOp GId G = ι u X | x X u G x = x
19 2 18 syl5eq G GrpOp U = ι u X | x X u G x = x