Metamath Proof Explorer


Theorem fmucndlem

Description: Lemma for fmucnd . (Contributed by Thierry Arnoux, 19-Nov-2017)

Ref Expression
Assertion fmucndlem ⊢ F Fn X ∧ A ⊆ X → x ∈ X , y ∈ X ⟼ F ⁡ x F ⁡ y A × A = F A × F A

Proof

Step Hyp Ref Expression
1 df-ima ⊢ x ∈ X , y ∈ X ⟼ F ⁡ x F ⁡ y A × A = ran ⁡ x ∈ X , y ∈ X ⟼ F ⁡ x F ⁡ y ↾ A × A
2 simpr ⊢ F Fn X ∧ A ⊆ X → A ⊆ X
3 resmpo ⊢ A ⊆ X ∧ A ⊆ X → x ∈ X , y ∈ X ⟼ F ⁡ x F ⁡ y ↾ A × A = x ∈ A , y ∈ A ⟼ F ⁡ x F ⁡ y
4 2 3 sylancom ⊢ F Fn X ∧ A ⊆ X → x ∈ X , y ∈ X ⟼ F ⁡ x F ⁡ y ↾ A × A = x ∈ A , y ∈ A ⟼ F ⁡ x F ⁡ y
5 4 rneqd ⊢ F Fn X ∧ A ⊆ X → ran ⁡ x ∈ X , y ∈ X ⟼ F ⁡ x F ⁡ y ↾ A × A = ran ⁡ x ∈ A , y ∈ A ⟼ F ⁡ x F ⁡ y
6 1 5 eqtrid ⊢ F Fn X ∧ A ⊆ X → x ∈ X , y ∈ X ⟼ F ⁡ x F ⁡ y A × A = ran ⁡ x ∈ A , y ∈ A ⟼ F ⁡ x F ⁡ y
7 vex ⊢ x ∈ V
8 vex ⊢ y ∈ V
9 7 8 op1std ⊢ p = x y → 1 st ⁡ p = x
10 9 fveq2d ⊢ p = x y → F ⁡ 1 st ⁡ p = F ⁡ x
11 7 8 op2ndd ⊢ p = x y → 2 nd ⁡ p = y
12 11 fveq2d ⊢ p = x y → F ⁡ 2 nd ⁡ p = F ⁡ y
13 10 12 opeq12d ⊢ p = x y → F ⁡ 1 st ⁡ p F ⁡ 2 nd ⁡ p = F ⁡ x F ⁡ y
14 13 mpompt ⊢ p ∈ A × A ⟼ F ⁡ 1 st ⁡ p F ⁡ 2 nd ⁡ p = x ∈ A , y ∈ A ⟼ F ⁡ x F ⁡ y
15 14 eqcomi ⊢ x ∈ A , y ∈ A ⟼ F ⁡ x F ⁡ y = p ∈ A × A ⟼ F ⁡ 1 st ⁡ p F ⁡ 2 nd ⁡ p
16 15 rneqi ⊢ ran ⁡ x ∈ A , y ∈ A ⟼ F ⁡ x F ⁡ y = ran ⁡ p ∈ A × A ⟼ F ⁡ 1 st ⁡ p F ⁡ 2 nd ⁡ p
17 fvexd ⊢ ⊤ ∧ p ∈ A × A → F ⁡ 1 st ⁡ p ∈ V
18 fvexd ⊢ ⊤ ∧ p ∈ A × A → F ⁡ 2 nd ⁡ p ∈ V
19 16 17 18 fliftrel ⊢ ⊤ → ran ⁡ x ∈ A , y ∈ A ⟼ F ⁡ x F ⁡ y ⊆ V × V
20 19 mptru ⊢ ran ⁡ x ∈ A , y ∈ A ⟼ F ⁡ x F ⁡ y ⊆ V × V
21 20 sseli ⊢ p ∈ ran ⁡ x ∈ A , y ∈ A ⟼ F ⁡ x F ⁡ y → p ∈ V × V
22 21 adantl ⊢ F Fn X ∧ A ⊆ X ∧ p ∈ ran ⁡ x ∈ A , y ∈ A ⟼ F ⁡ x F ⁡ y → p ∈ V × V
23 xpss ⊢ F A × F A ⊆ V × V
24 23 sseli ⊢ p ∈ F A × F A → p ∈ V × V
25 24 adantl ⊢ F Fn X ∧ A ⊆ X ∧ p ∈ F A × F A → p ∈ V × V
26 eqid ⊢ x ∈ A , y ∈ A ⟼ F ⁡ x F ⁡ y = x ∈ A , y ∈ A ⟼ F ⁡ x F ⁡ y
27 opex ⊢ F ⁡ x F ⁡ y ∈ V
28 26 27 elrnmpo ⊢ 1 st ⁡ p 2 nd ⁡ p ∈ ran ⁡ x ∈ A , y ∈ A ⟼ F ⁡ x F ⁡ y ↔ ∃ x ∈ A ∃ y ∈ A 1 st ⁡ p 2 nd ⁡ p = F ⁡ x F ⁡ y
29 eqcom ⊢ 1 st ⁡ p 2 nd ⁡ p = F ⁡ x F ⁡ y ↔ F ⁡ x F ⁡ y = 1 st ⁡ p 2 nd ⁡ p
30 fvex ⊢ 1 st ⁡ p ∈ V
31 fvex ⊢ 2 nd ⁡ p ∈ V
32 30 31 opth2 ⊢ F ⁡ x F ⁡ y = 1 st ⁡ p 2 nd ⁡ p ↔ F ⁡ x = 1 st ⁡ p ∧ F ⁡ y = 2 nd ⁡ p
33 29 32 bitri ⊢ 1 st ⁡ p 2 nd ⁡ p = F ⁡ x F ⁡ y ↔ F ⁡ x = 1 st ⁡ p ∧ F ⁡ y = 2 nd ⁡ p
34 33 2rexbii ⊢ ∃ x ∈ A ∃ y ∈ A 1 st ⁡ p 2 nd ⁡ p = F ⁡ x F ⁡ y ↔ ∃ x ∈ A ∃ y ∈ A F ⁡ x = 1 st ⁡ p ∧ F ⁡ y = 2 nd ⁡ p
35 reeanv ⊢ ∃ x ∈ A ∃ y ∈ A F ⁡ x = 1 st ⁡ p ∧ F ⁡ y = 2 nd ⁡ p ↔ ∃ x ∈ A F ⁡ x = 1 st ⁡ p ∧ ∃ y ∈ A F ⁡ y = 2 nd ⁡ p
36 28 34 35 3bitri ⊢ 1 st ⁡ p 2 nd ⁡ p ∈ ran ⁡ x ∈ A , y ∈ A ⟼ F ⁡ x F ⁡ y ↔ ∃ x ∈ A F ⁡ x = 1 st ⁡ p ∧ ∃ y ∈ A F ⁡ y = 2 nd ⁡ p
37 fvelimab ⊢ F Fn X ∧ A ⊆ X → 1 st ⁡ p ∈ F A ↔ ∃ x ∈ A F ⁡ x = 1 st ⁡ p
38 fvelimab ⊢ F Fn X ∧ A ⊆ X → 2 nd ⁡ p ∈ F A ↔ ∃ y ∈ A F ⁡ y = 2 nd ⁡ p
39 37 38 anbi12d ⊢ F Fn X ∧ A ⊆ X → 1 st ⁡ p ∈ F A ∧ 2 nd ⁡ p ∈ F A ↔ ∃ x ∈ A F ⁡ x = 1 st ⁡ p ∧ ∃ y ∈ A F ⁡ y = 2 nd ⁡ p
40 36 39 bitr4id ⊢ F Fn X ∧ A ⊆ X → 1 st ⁡ p 2 nd ⁡ p ∈ ran ⁡ x ∈ A , y ∈ A ⟼ F ⁡ x F ⁡ y ↔ 1 st ⁡ p ∈ F A ∧ 2 nd ⁡ p ∈ F A
41 opelxp ⊢ 1 st ⁡ p 2 nd ⁡ p ∈ F A × F A ↔ 1 st ⁡ p ∈ F A ∧ 2 nd ⁡ p ∈ F A
42 40 41 bitr4di ⊢ F Fn X ∧ A ⊆ X → 1 st ⁡ p 2 nd ⁡ p ∈ ran ⁡ x ∈ A , y ∈ A ⟼ F ⁡ x F ⁡ y ↔ 1 st ⁡ p 2 nd ⁡ p ∈ F A × F A
43 42 adantr ⊢ F Fn X ∧ A ⊆ X ∧ p ∈ V × V → 1 st ⁡ p 2 nd ⁡ p ∈ ran ⁡ x ∈ A , y ∈ A ⟼ F ⁡ x F ⁡ y ↔ 1 st ⁡ p 2 nd ⁡ p ∈ F A × F A
44 1st2nd2 ⊢ p ∈ V × V → p = 1 st ⁡ p 2 nd ⁡ p
45 44 adantl ⊢ F Fn X ∧ A ⊆ X ∧ p ∈ V × V → p = 1 st ⁡ p 2 nd ⁡ p
46 45 eleq1d ⊢ F Fn X ∧ A ⊆ X ∧ p ∈ V × V → p ∈ ran ⁡ x ∈ A , y ∈ A ⟼ F ⁡ x F ⁡ y ↔ 1 st ⁡ p 2 nd ⁡ p ∈ ran ⁡ x ∈ A , y ∈ A ⟼ F ⁡ x F ⁡ y
47 45 eleq1d ⊢ F Fn X ∧ A ⊆ X ∧ p ∈ V × V → p ∈ F A × F A ↔ 1 st ⁡ p 2 nd ⁡ p ∈ F A × F A
48 43 46 47 3bitr4d ⊢ F Fn X ∧ A ⊆ X ∧ p ∈ V × V → p ∈ ran ⁡ x ∈ A , y ∈ A ⟼ F ⁡ x F ⁡ y ↔ p ∈ F A × F A
49 22 25 48 eqrdav ⊢ F Fn X ∧ A ⊆ X → ran ⁡ x ∈ A , y ∈ A ⟼ F ⁡ x F ⁡ y = F A × F A
50 6 49 eqtrd ⊢ F Fn X ∧ A ⊆ X → x ∈ X , y ∈ X ⟼ F ⁡ x F ⁡ y A × A = F A × F A