Metamath Proof Explorer


Theorem sbthlem2

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
Assertion sbthlem2 ran g A A g B f D D

Proof

Step Hyp Ref Expression
1 sbthlem.1 A V
2 sbthlem.2 D = x | x A g B f x A x
3 1 2 sbthlem1 D A g B f D
4 imass2 D A g B f D f D f A g B f D
5 sscon f D f A g B f D B f A g B f D B f D
6 3 4 5 mp2b B f A g B f D B f D
7 imass2 B f A g B f D B f D g B f A g B f D g B f D
8 sscon g B f A g B f D g B f D A g B f D A g B f A g B f D
9 6 7 8 mp2b A g B f D A g B f A g B f D
10 imassrn g B f A g B f D ran g
11 sstr2 g B f A g B f D ran g ran g A g B f A g B f D A
12 10 11 ax-mp ran g A g B f A g B f D A
13 difss A g B f D A
14 ssconb g B f A g B f D A A g B f D A g B f A g B f D A A g B f D A g B f D A g B f A g B f D
15 12 13 14 sylancl ran g A g B f A g B f D A A g B f D A g B f D A g B f A g B f D
16 9 15 mpbiri ran g A g B f A g B f D A A g B f D
17 16 13 jctil ran g A A g B f D A g B f A g B f D A A g B f D
18 1 difexi A g B f D V
19 sseq1 x = A g B f D x A A g B f D A
20 imaeq2 x = A g B f D f x = f A g B f D
21 20 difeq2d x = A g B f D B f x = B f A g B f D
22 21 imaeq2d x = A g B f D g B f x = g B f A g B f D
23 difeq2 x = A g B f D A x = A A g B f D
24 22 23 sseq12d x = A g B f D g B f x A x g B f A g B f D A A g B f D
25 19 24 anbi12d x = A g B f D x A g B f x A x A g B f D A g B f A g B f D A A g B f D
26 18 25 elab A g B f D x | x A g B f x A x A g B f D A g B f A g B f D A A g B f D
27 17 26 sylibr ran g A A g B f D x | x A g B f x A x
28 27 2 eleqtrrdi ran g A A g B f D D
29 elssuni A g B f D D A g B f D D
30 28 29 syl ran g A A g B f D D