Metamath Proof Explorer


Theorem xpsff1o

Description: The function appearing in xpsval is a bijection from the cartesian product to the indexed cartesian product indexed on the pair 2o = { (/) , 1o } . (Contributed by Mario Carneiro, 15-Aug-2015)

Ref Expression
Hypothesis xpsff1o.f ⊢ F = x ∈ A , y ∈ B ⟼ ∅ x 1 𝑜 y
Assertion xpsff1o ⊢ F : A × B ⟶ 1-1 onto ⨉ k ∈ 2 𝑜 if k = ∅ A B

Proof

Step Hyp Ref Expression
1 xpsff1o.f ⊢ F = x ∈ A , y ∈ B ⟼ ∅ x 1 𝑜 y
2 xpsfrnel2 ⊢ ∅ x 1 𝑜 y ∈ ⨉ k ∈ 2 𝑜 if k = ∅ A B ↔ x ∈ A ∧ y ∈ B
3 2 biimpri ⊢ x ∈ A ∧ y ∈ B → ∅ x 1 𝑜 y ∈ ⨉ k ∈ 2 𝑜 if k = ∅ A B
4 3 rgen2 ⊢ ∀ x ∈ A ∀ y ∈ B ∅ x 1 𝑜 y ∈ ⨉ k ∈ 2 𝑜 if k = ∅ A B
5 1 fmpo ⊢ ∀ x ∈ A ∀ y ∈ B ∅ x 1 𝑜 y ∈ ⨉ k ∈ 2 𝑜 if k = ∅ A B ↔ F : A × B ⟶ ⨉ k ∈ 2 𝑜 if k = ∅ A B
6 4 5 mpbi ⊢ F : A × B ⟶ ⨉ k ∈ 2 𝑜 if k = ∅ A B
7 1st2nd2 ⊢ z ∈ A × B → z = 1 st ⁡ z 2 nd ⁡ z
8 7 fveq2d ⊢ z ∈ A × B → F ⁡ z = F ⁡ 1 st ⁡ z 2 nd ⁡ z
9 df-ov ⊢ 1 st ⁡ z F 2 nd ⁡ z = F ⁡ 1 st ⁡ z 2 nd ⁡ z
10 xp1st ⊢ z ∈ A × B → 1 st ⁡ z ∈ A
11 xp2nd ⊢ z ∈ A × B → 2 nd ⁡ z ∈ B
12 1 xpsfval ⊢ 1 st ⁡ z ∈ A ∧ 2 nd ⁡ z ∈ B → 1 st ⁡ z F 2 nd ⁡ z = ∅ 1 st ⁡ z 1 𝑜 2 nd ⁡ z
13 10 11 12 syl2anc ⊢ z ∈ A × B → 1 st ⁡ z F 2 nd ⁡ z = ∅ 1 st ⁡ z 1 𝑜 2 nd ⁡ z
14 9 13 eqtr3id ⊢ z ∈ A × B → F ⁡ 1 st ⁡ z 2 nd ⁡ z = ∅ 1 st ⁡ z 1 𝑜 2 nd ⁡ z
15 8 14 eqtrd ⊢ z ∈ A × B → F ⁡ z = ∅ 1 st ⁡ z 1 𝑜 2 nd ⁡ z
16 1st2nd2 ⊢ w ∈ A × B → w = 1 st ⁡ w 2 nd ⁡ w
17 16 fveq2d ⊢ w ∈ A × B → F ⁡ w = F ⁡ 1 st ⁡ w 2 nd ⁡ w
18 df-ov ⊢ 1 st ⁡ w F 2 nd ⁡ w = F ⁡ 1 st ⁡ w 2 nd ⁡ w
19 xp1st ⊢ w ∈ A × B → 1 st ⁡ w ∈ A
20 xp2nd ⊢ w ∈ A × B → 2 nd ⁡ w ∈ B
21 1 xpsfval ⊢ 1 st ⁡ w ∈ A ∧ 2 nd ⁡ w ∈ B → 1 st ⁡ w F 2 nd ⁡ w = ∅ 1 st ⁡ w 1 𝑜 2 nd ⁡ w
22 19 20 21 syl2anc ⊢ w ∈ A × B → 1 st ⁡ w F 2 nd ⁡ w = ∅ 1 st ⁡ w 1 𝑜 2 nd ⁡ w
23 18 22 eqtr3id ⊢ w ∈ A × B → F ⁡ 1 st ⁡ w 2 nd ⁡ w = ∅ 1 st ⁡ w 1 𝑜 2 nd ⁡ w
24 17 23 eqtrd ⊢ w ∈ A × B → F ⁡ w = ∅ 1 st ⁡ w 1 𝑜 2 nd ⁡ w
25 15 24 eqeqan12d ⊢ z ∈ A × B ∧ w ∈ A × B → F ⁡ z = F ⁡ w ↔ ∅ 1 st ⁡ z 1 𝑜 2 nd ⁡ z = ∅ 1 st ⁡ w 1 𝑜 2 nd ⁡ w
26 fveq1 ⊢ ∅ 1 st ⁡ z 1 𝑜 2 nd ⁡ z = ∅ 1 st ⁡ w 1 𝑜 2 nd ⁡ w → ∅ 1 st ⁡ z 1 𝑜 2 nd ⁡ z ⁡ ∅ = ∅ 1 st ⁡ w 1 𝑜 2 nd ⁡ w ⁡ ∅
27 fvex ⊢ 1 st ⁡ z ∈ V
28 fvpr0o ⊢ 1 st ⁡ z ∈ V → ∅ 1 st ⁡ z 1 𝑜 2 nd ⁡ z ⁡ ∅ = 1 st ⁡ z
29 27 28 ax-mp ⊢ ∅ 1 st ⁡ z 1 𝑜 2 nd ⁡ z ⁡ ∅ = 1 st ⁡ z
30 fvex ⊢ 1 st ⁡ w ∈ V
31 fvpr0o ⊢ 1 st ⁡ w ∈ V → ∅ 1 st ⁡ w 1 𝑜 2 nd ⁡ w ⁡ ∅ = 1 st ⁡ w
32 30 31 ax-mp ⊢ ∅ 1 st ⁡ w 1 𝑜 2 nd ⁡ w ⁡ ∅ = 1 st ⁡ w
33 26 29 32 3eqtr3g ⊢ ∅ 1 st ⁡ z 1 𝑜 2 nd ⁡ z = ∅ 1 st ⁡ w 1 𝑜 2 nd ⁡ w → 1 st ⁡ z = 1 st ⁡ w
34 fveq1 ⊢ ∅ 1 st ⁡ z 1 𝑜 2 nd ⁡ z = ∅ 1 st ⁡ w 1 𝑜 2 nd ⁡ w → ∅ 1 st ⁡ z 1 𝑜 2 nd ⁡ z ⁡ 1 𝑜 = ∅ 1 st ⁡ w 1 𝑜 2 nd ⁡ w ⁡ 1 𝑜
35 fvex ⊢ 2 nd ⁡ z ∈ V
36 fvpr1o ⊢ 2 nd ⁡ z ∈ V → ∅ 1 st ⁡ z 1 𝑜 2 nd ⁡ z ⁡ 1 𝑜 = 2 nd ⁡ z
37 35 36 ax-mp ⊢ ∅ 1 st ⁡ z 1 𝑜 2 nd ⁡ z ⁡ 1 𝑜 = 2 nd ⁡ z
38 fvex ⊢ 2 nd ⁡ w ∈ V
39 fvpr1o ⊢ 2 nd ⁡ w ∈ V → ∅ 1 st ⁡ w 1 𝑜 2 nd ⁡ w ⁡ 1 𝑜 = 2 nd ⁡ w
40 38 39 ax-mp ⊢ ∅ 1 st ⁡ w 1 𝑜 2 nd ⁡ w ⁡ 1 𝑜 = 2 nd ⁡ w
41 34 37 40 3eqtr3g ⊢ ∅ 1 st ⁡ z 1 𝑜 2 nd ⁡ z = ∅ 1 st ⁡ w 1 𝑜 2 nd ⁡ w → 2 nd ⁡ z = 2 nd ⁡ w
42 33 41 opeq12d ⊢ ∅ 1 st ⁡ z 1 𝑜 2 nd ⁡ z = ∅ 1 st ⁡ w 1 𝑜 2 nd ⁡ w → 1 st ⁡ z 2 nd ⁡ z = 1 st ⁡ w 2 nd ⁡ w
43 7 16 eqeqan12d ⊢ z ∈ A × B ∧ w ∈ A × B → z = w ↔ 1 st ⁡ z 2 nd ⁡ z = 1 st ⁡ w 2 nd ⁡ w
44 42 43 imbitrrid ⊢ z ∈ A × B ∧ w ∈ A × B → ∅ 1 st ⁡ z 1 𝑜 2 nd ⁡ z = ∅ 1 st ⁡ w 1 𝑜 2 nd ⁡ w → z = w
45 25 44 sylbid ⊢ z ∈ A × B ∧ w ∈ A × B → F ⁡ z = F ⁡ w → z = w
46 45 rgen2 ⊢ ∀ z ∈ A × B ∀ w ∈ A × B F ⁡ z = F ⁡ w → z = w
47 dff13 ⊢ F : A × B ⟶ 1-1 ⨉ k ∈ 2 𝑜 if k = ∅ A B ↔ F : A × B ⟶ ⨉ k ∈ 2 𝑜 if k = ∅ A B ∧ ∀ z ∈ A × B ∀ w ∈ A × B F ⁡ z = F ⁡ w → z = w
48 6 46 47 mpbir2an ⊢ F : A × B ⟶ 1-1 ⨉ k ∈ 2 𝑜 if k = ∅ A B
49 xpsfrnel ⊢ z ∈ ⨉ k ∈ 2 𝑜 if k = ∅ A B ↔ z Fn 2 𝑜 ∧ z ⁡ ∅ ∈ A ∧ z ⁡ 1 𝑜 ∈ B
50 49 simp2bi ⊢ z ∈ ⨉ k ∈ 2 𝑜 if k = ∅ A B → z ⁡ ∅ ∈ A
51 49 simp3bi ⊢ z ∈ ⨉ k ∈ 2 𝑜 if k = ∅ A B → z ⁡ 1 𝑜 ∈ B
52 1 xpsfval ⊢ z ⁡ ∅ ∈ A ∧ z ⁡ 1 𝑜 ∈ B → z ⁡ ∅ F z ⁡ 1 𝑜 = ∅ z ⁡ ∅ 1 𝑜 z ⁡ 1 𝑜
53 50 51 52 syl2anc ⊢ z ∈ ⨉ k ∈ 2 𝑜 if k = ∅ A B → z ⁡ ∅ F z ⁡ 1 𝑜 = ∅ z ⁡ ∅ 1 𝑜 z ⁡ 1 𝑜
54 ixpfn ⊢ z ∈ ⨉ k ∈ 2 𝑜 if k = ∅ A B → z Fn 2 𝑜
55 xpsfeq ⊢ z Fn 2 𝑜 → ∅ z ⁡ ∅ 1 𝑜 z ⁡ 1 𝑜 = z
56 54 55 syl ⊢ z ∈ ⨉ k ∈ 2 𝑜 if k = ∅ A B → ∅ z ⁡ ∅ 1 𝑜 z ⁡ 1 𝑜 = z
57 53 56 eqtr2d ⊢ z ∈ ⨉ k ∈ 2 𝑜 if k = ∅ A B → z = z ⁡ ∅ F z ⁡ 1 𝑜
58 rspceov ⊢ z ⁡ ∅ ∈ A ∧ z ⁡ 1 𝑜 ∈ B ∧ z = z ⁡ ∅ F z ⁡ 1 𝑜 → ∃ a ∈ A ∃ b ∈ B z = a F b
59 50 51 57 58 syl3anc ⊢ z ∈ ⨉ k ∈ 2 𝑜 if k = ∅ A B → ∃ a ∈ A ∃ b ∈ B z = a F b
60 59 rgen ⊢ ∀ z ∈ ⨉ k ∈ 2 𝑜 if k = ∅ A B ∃ a ∈ A ∃ b ∈ B z = a F b
61 foov ⊢ F : A × B ⟶ onto ⨉ k ∈ 2 𝑜 if k = ∅ A B ↔ F : A × B ⟶ ⨉ k ∈ 2 𝑜 if k = ∅ A B ∧ ∀ z ∈ ⨉ k ∈ 2 𝑜 if k = ∅ A B ∃ a ∈ A ∃ b ∈ B z = a F b
62 6 60 61 mpbir2an ⊢ F : A × B ⟶ onto ⨉ k ∈ 2 𝑜 if k = ∅ A B
63 df-f1o ⊢ F : A × B ⟶ 1-1 onto ⨉ k ∈ 2 𝑜 if k = ∅ A B ↔ F : A × B ⟶ 1-1 ⨉ k ∈ 2 𝑜 if k = ∅ A B ∧ F : A × B ⟶ onto ⨉ k ∈ 2 𝑜 if k = ∅ A B
64 48 62 63 mpbir2an ⊢ F : A × B ⟶ 1-1 onto ⨉ k ∈ 2 𝑜 if k = ∅ A B