Metamath Proof Explorer


Theorem sbthlem5

Description: Lemma for sbth . (Contributed by NM, 22-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 sbthlem5 dom f = A ran g A dom H = A

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 1 2 sbthlem1 D A g B f D
5 difss A g B f D A
6 4 5 sstri D A
7 sseq2 dom f = A D dom f D A
8 6 7 mpbiri dom f = A D dom f
9 dfss D dom f D = D dom f
10 8 9 sylib dom f = A D = D dom f
11 10 uneq1d dom f = A D A D = D dom f A D
12 1 2 sbthlem3 ran g A g B f D = A D
13 imassrn g B f D ran g
14 12 13 eqsstrrdi ran g A A D ran g
15 dfss A D ran g A D = A D ran g
16 14 15 sylib ran g A A D = A D ran g
17 16 uneq2d ran g A D dom f A D = D dom f A D ran g
18 11 17 sylan9eq dom f = A ran g A D A D = D dom f A D ran g
19 3 dmeqi dom H = dom f D g -1 A D
20 dmun dom f D g -1 A D = dom f D dom g -1 A D
21 dmres dom f D = D dom f
22 dmres dom g -1 A D = A D dom g -1
23 df-rn ran g = dom g -1
24 23 eqcomi dom g -1 = ran g
25 24 ineq2i A D dom g -1 = A D ran g
26 22 25 eqtri dom g -1 A D = A D ran g
27 21 26 uneq12i dom f D dom g -1 A D = D dom f A D ran g
28 20 27 eqtri dom f D g -1 A D = D dom f A D ran g
29 19 28 eqtri dom H = D dom f A D ran g
30 18 29 syl6reqr dom f = A ran g A dom H = D A D
31 undif D A D A D = A
32 6 31 mpbi D A D = A
33 30 32 syl6eq dom f = A ran g A dom H = A