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 3 dmeqi dom H = dom f D g -1 A D
5 dmun dom f D g -1 A D = dom f D dom g -1 A D
6 dmres dom f D = D dom f
7 dmres dom g -1 A D = A D dom g -1
8 df-rn ran g = dom g -1
9 8 eqcomi dom g -1 = ran g
10 9 ineq2i A D dom g -1 = A D ran g
11 7 10 eqtri dom g -1 A D = A D ran g
12 6 11 uneq12i dom f D dom g -1 A D = D dom f A D ran g
13 5 12 eqtri dom f D g -1 A D = D dom f A D ran g
14 4 13 eqtri dom H = D dom f A D ran g
15 1 2 sbthlem1 D A g B f D
16 difss A g B f D A
17 15 16 sstri D A
18 sseq2 dom f = A D dom f D A
19 17 18 mpbiri dom f = A D dom f
20 dfss D dom f D = D dom f
21 19 20 sylib dom f = A D = D dom f
22 21 uneq1d dom f = A D A D = D dom f A D
23 1 2 sbthlem3 ran g A g B f D = A D
24 imassrn g B f D ran g
25 23 24 eqsstrrdi ran g A A D ran g
26 dfss A D ran g A D = A D ran g
27 25 26 sylib ran g A A D = A D ran g
28 27 uneq2d ran g A D dom f A D = D dom f A D ran g
29 22 28 sylan9eq dom f = A ran g A D A D = D dom f A D ran g
30 14 29 eqtr4id dom f = A ran g A dom H = D A D
31 undif D A D A D = A
32 17 31 mpbi D A D = A
33 30 32 eqtrdi dom f = A ran g A dom H = A