Metamath Proof Explorer


Theorem xpinpreima2

Description: Rewrite the cartesian product of two sets as the intersection of their preimage by 1st and 2nd , the projections on the first and second elements. (Contributed by Thierry Arnoux, 22-Sep-2017)

Ref Expression
Assertion xpinpreima2 ⊢ A ⊆ E ∧ B ⊆ F → A × B = 1 st ↾ E × F -1 A ∩ 2 nd ↾ E × F -1 B

Proof

Step Hyp Ref Expression
1 xp2 ⊢ A × B = r ∈ V × V | 1 st ⁡ r ∈ A ∧ 2 nd ⁡ r ∈ B
2 xpss ⊢ E × F ⊆ V × V
3 rabss2 ⊢ E × F ⊆ V × V → r ∈ E × F | 1 st ⁡ r ∈ A ∧ 2 nd ⁡ r ∈ B ⊆ r ∈ V × V | 1 st ⁡ r ∈ A ∧ 2 nd ⁡ r ∈ B
4 2 3 mp1i ⊢ A ⊆ E ∧ B ⊆ F → r ∈ E × F | 1 st ⁡ r ∈ A ∧ 2 nd ⁡ r ∈ B ⊆ r ∈ V × V | 1 st ⁡ r ∈ A ∧ 2 nd ⁡ r ∈ B
5 simprl ⊢ A ⊆ E ∧ B ⊆ F ∧ r ∈ V × V ∧ 1 st ⁡ r ∈ A ∧ 2 nd ⁡ r ∈ B → r ∈ V × V
6 simpll ⊢ A ⊆ E ∧ B ⊆ F ∧ r ∈ V × V ∧ 1 st ⁡ r ∈ A ∧ 2 nd ⁡ r ∈ B → A ⊆ E
7 simprrl ⊢ A ⊆ E ∧ B ⊆ F ∧ r ∈ V × V ∧ 1 st ⁡ r ∈ A ∧ 2 nd ⁡ r ∈ B → 1 st ⁡ r ∈ A
8 6 7 sseldd ⊢ A ⊆ E ∧ B ⊆ F ∧ r ∈ V × V ∧ 1 st ⁡ r ∈ A ∧ 2 nd ⁡ r ∈ B → 1 st ⁡ r ∈ E
9 simplr ⊢ A ⊆ E ∧ B ⊆ F ∧ r ∈ V × V ∧ 1 st ⁡ r ∈ A ∧ 2 nd ⁡ r ∈ B → B ⊆ F
10 simprrr ⊢ A ⊆ E ∧ B ⊆ F ∧ r ∈ V × V ∧ 1 st ⁡ r ∈ A ∧ 2 nd ⁡ r ∈ B → 2 nd ⁡ r ∈ B
11 9 10 sseldd ⊢ A ⊆ E ∧ B ⊆ F ∧ r ∈ V × V ∧ 1 st ⁡ r ∈ A ∧ 2 nd ⁡ r ∈ B → 2 nd ⁡ r ∈ F
12 8 11 jca ⊢ A ⊆ E ∧ B ⊆ F ∧ r ∈ V × V ∧ 1 st ⁡ r ∈ A ∧ 2 nd ⁡ r ∈ B → 1 st ⁡ r ∈ E ∧ 2 nd ⁡ r ∈ F
13 elxp7 ⊢ r ∈ E × F ↔ r ∈ V × V ∧ 1 st ⁡ r ∈ E ∧ 2 nd ⁡ r ∈ F
14 5 12 13 sylanbrc ⊢ A ⊆ E ∧ B ⊆ F ∧ r ∈ V × V ∧ 1 st ⁡ r ∈ A ∧ 2 nd ⁡ r ∈ B → r ∈ E × F
15 14 rabss3d ⊢ A ⊆ E ∧ B ⊆ F → r ∈ V × V | 1 st ⁡ r ∈ A ∧ 2 nd ⁡ r ∈ B ⊆ r ∈ E × F | 1 st ⁡ r ∈ A ∧ 2 nd ⁡ r ∈ B
16 4 15 eqssd ⊢ A ⊆ E ∧ B ⊆ F → r ∈ E × F | 1 st ⁡ r ∈ A ∧ 2 nd ⁡ r ∈ B = r ∈ V × V | 1 st ⁡ r ∈ A ∧ 2 nd ⁡ r ∈ B
17 1 16 eqtr4id ⊢ A ⊆ E ∧ B ⊆ F → A × B = r ∈ E × F | 1 st ⁡ r ∈ A ∧ 2 nd ⁡ r ∈ B
18 inrab ⊢ r ∈ E × F | 1 st ⁡ r ∈ A ∩ r ∈ E × F | 2 nd ⁡ r ∈ B = r ∈ E × F | 1 st ⁡ r ∈ A ∧ 2 nd ⁡ r ∈ B
19 17 18 eqtr4di ⊢ A ⊆ E ∧ B ⊆ F → A × B = r ∈ E × F | 1 st ⁡ r ∈ A ∩ r ∈ E × F | 2 nd ⁡ r ∈ B
20 f1stres ⊢ 1 st ↾ E × F : E × F ⟶ E
21 ffn ⊢ 1 st ↾ E × F : E × F ⟶ E → 1 st ↾ E × F Fn E × F
22 fncnvima2 ⊢ 1 st ↾ E × F Fn E × F → 1 st ↾ E × F -1 A = r ∈ E × F | 1 st ↾ E × F ⁡ r ∈ A
23 20 21 22 mp2b ⊢ 1 st ↾ E × F -1 A = r ∈ E × F | 1 st ↾ E × F ⁡ r ∈ A
24 fvres ⊢ r ∈ E × F → 1 st ↾ E × F ⁡ r = 1 st ⁡ r
25 24 eleq1d ⊢ r ∈ E × F → 1 st ↾ E × F ⁡ r ∈ A ↔ 1 st ⁡ r ∈ A
26 25 rabbiia ⊢ r ∈ E × F | 1 st ↾ E × F ⁡ r ∈ A = r ∈ E × F | 1 st ⁡ r ∈ A
27 23 26 eqtri ⊢ 1 st ↾ E × F -1 A = r ∈ E × F | 1 st ⁡ r ∈ A
28 f2ndres ⊢ 2 nd ↾ E × F : E × F ⟶ F
29 ffn ⊢ 2 nd ↾ E × F : E × F ⟶ F → 2 nd ↾ E × F Fn E × F
30 fncnvima2 ⊢ 2 nd ↾ E × F Fn E × F → 2 nd ↾ E × F -1 B = r ∈ E × F | 2 nd ↾ E × F ⁡ r ∈ B
31 28 29 30 mp2b ⊢ 2 nd ↾ E × F -1 B = r ∈ E × F | 2 nd ↾ E × F ⁡ r ∈ B
32 fvres ⊢ r ∈ E × F → 2 nd ↾ E × F ⁡ r = 2 nd ⁡ r
33 32 eleq1d ⊢ r ∈ E × F → 2 nd ↾ E × F ⁡ r ∈ B ↔ 2 nd ⁡ r ∈ B
34 33 rabbiia ⊢ r ∈ E × F | 2 nd ↾ E × F ⁡ r ∈ B = r ∈ E × F | 2 nd ⁡ r ∈ B
35 31 34 eqtri ⊢ 2 nd ↾ E × F -1 B = r ∈ E × F | 2 nd ⁡ r ∈ B
36 27 35 ineq12i ⊢ 1 st ↾ E × F -1 A ∩ 2 nd ↾ E × F -1 B = r ∈ E × F | 1 st ⁡ r ∈ A ∩ r ∈ E × F | 2 nd ⁡ r ∈ B
37 19 36 eqtr4di ⊢ A ⊆ E ∧ B ⊆ F → A × B = 1 st ↾ E × F -1 A ∩ 2 nd ↾ E × F -1 B