Metamath Proof Explorer


Theorem dprdf1

Description: Rearrange the index set of a direct product family. (Contributed by Mario Carneiro, 25-Apr-2016)

Ref Expression
Hypotheses dprdf1.1
|- ( ph -> G dom DProd S )
dprdf1.2
|- ( ph -> dom S = I )
dprdf1.3
|- ( ph -> F : J -1-1-> I )
Assertion dprdf1
|- ( ph -> ( G dom DProd ( S o. F ) /\ ( G DProd ( S o. F ) ) C_ ( G DProd S ) ) )

Proof

Step Hyp Ref Expression
1 dprdf1.1
 |-  ( ph -> G dom DProd S )
2 dprdf1.2
 |-  ( ph -> dom S = I )
3 dprdf1.3
 |-  ( ph -> F : J -1-1-> I )
4 f1f
 |-  ( F : J -1-1-> I -> F : J --> I )
5 frn
 |-  ( F : J --> I -> ran F C_ I )
6 3 4 5 3syl
 |-  ( ph -> ran F C_ I )
7 1 2 6 dprdres
 |-  ( ph -> ( G dom DProd ( S |` ran F ) /\ ( G DProd ( S |` ran F ) ) C_ ( G DProd S ) ) )
8 7 simpld
 |-  ( ph -> G dom DProd ( S |` ran F ) )
9 1 2 dprdf2
 |-  ( ph -> S : I --> ( SubGrp ` G ) )
10 9 6 fssresd
 |-  ( ph -> ( S |` ran F ) : ran F --> ( SubGrp ` G ) )
11 10 fdmd
 |-  ( ph -> dom ( S |` ran F ) = ran F )
12 f1f1orn
 |-  ( F : J -1-1-> I -> F : J -1-1-onto-> ran F )
13 3 12 syl
 |-  ( ph -> F : J -1-1-onto-> ran F )
14 8 11 13 dprdf1o
 |-  ( ph -> ( G dom DProd ( ( S |` ran F ) o. F ) /\ ( G DProd ( ( S |` ran F ) o. F ) ) = ( G DProd ( S |` ran F ) ) ) )
15 14 simpld
 |-  ( ph -> G dom DProd ( ( S |` ran F ) o. F ) )
16 ssid
 |-  ran F C_ ran F
17 cores
 |-  ( ran F C_ ran F -> ( ( S |` ran F ) o. F ) = ( S o. F ) )
18 16 17 ax-mp
 |-  ( ( S |` ran F ) o. F ) = ( S o. F )
19 15 18 breqtrdi
 |-  ( ph -> G dom DProd ( S o. F ) )
20 18 oveq2i
 |-  ( G DProd ( ( S |` ran F ) o. F ) ) = ( G DProd ( S o. F ) )
21 14 simprd
 |-  ( ph -> ( G DProd ( ( S |` ran F ) o. F ) ) = ( G DProd ( S |` ran F ) ) )
22 20 21 eqtr3id
 |-  ( ph -> ( G DProd ( S o. F ) ) = ( G DProd ( S |` ran F ) ) )
23 7 simprd
 |-  ( ph -> ( G DProd ( S |` ran F ) ) C_ ( G DProd S ) )
24 22 23 eqsstrd
 |-  ( ph -> ( G DProd ( S o. F ) ) C_ ( G DProd S ) )
25 19 24 jca
 |-  ( ph -> ( G dom DProd ( S o. F ) /\ ( G DProd ( S o. F ) ) C_ ( G DProd S ) ) )