Metamath Proof Explorer


Theorem sbthlem6

Description: Lemma for sbth . (Contributed by NM, 27-Mar-1998)

Ref Expression
Hypotheses sbthlem.1 A V
sbthlem.2 D = x | x A g B f x A x
sbthlem.3 H = f D g -1 A D
Assertion sbthlem6 ran f B dom g = B ran g A Fun g -1 ran H = B

Proof

Step Hyp Ref Expression
1 sbthlem.1 A V
2 sbthlem.2 D = x | x A g B f x A x
3 sbthlem.3 H = f D g -1 A D
4 rnun ran f D g -1 A D = ran f D ran g -1 A D
5 3 rneqi ran H = ran f D g -1 A D
6 df-ima f D = ran f D
7 6 uneq1i f D ran g -1 A D = ran f D ran g -1 A D
8 4 5 7 3eqtr4i ran H = f D ran g -1 A D
9 df-ima g -1 A D = ran g -1 A D
10 1 2 sbthlem4 dom g = B ran g A Fun g -1 g -1 A D = B f D
11 9 10 syl5reqr dom g = B ran g A Fun g -1 B f D = ran g -1 A D
12 11 uneq2d dom g = B ran g A Fun g -1 f D B f D = f D ran g -1 A D
13 8 12 eqtr4id dom g = B ran g A Fun g -1 ran H = f D B f D
14 imassrn f D ran f
15 sstr2 f D ran f ran f B f D B
16 14 15 ax-mp ran f B f D B
17 undif f D B f D B f D = B
18 16 17 sylib ran f B f D B f D = B
19 13 18 sylan9eqr ran f B dom g = B ran g A Fun g -1 ran H = B