Metamath Proof Explorer


Theorem disjprg

Description: A pair collection is disjoint iff the two sets in the family have empty intersection. (Contributed by Mario Carneiro, 14-Nov-2016)

Ref Expression
Hypotheses disjprg.1 ⊢ x = A → C = D
disjprg.2 ⊢ x = B → C = E
Assertion disjprg ⊢ A ∈ V ∧ B ∈ V ∧ A ≠ B → Disj x ∈ A B C ↔ D ∩ E = ∅

Proof

Step Hyp Ref Expression
1 disjprg.1 ⊢ x = A → C = D
2 disjprg.2 ⊢ x = B → C = E
3 eqeq1 ⊢ y = A → y = z ↔ A = z
4 nfcv ⊢ Ⅎ _ x A
5 nfcv ⊢ Ⅎ _ x D
6 4 5 1 csbhypf ⊢ y = A → ⦋ y / x⦌ C = D
7 6 ineq1d ⊢ y = A → ⦋ y / x⦌ C ∩ ⦋ z / x⦌ C = D ∩ ⦋ z / x⦌ C
8 7 eqeq1d ⊢ y = A → ⦋ y / x⦌ C ∩ ⦋ z / x⦌ C = ∅ ↔ D ∩ ⦋ z / x⦌ C = ∅
9 3 8 orbi12d ⊢ y = A → y = z ∨ ⦋ y / x⦌ C ∩ ⦋ z / x⦌ C = ∅ ↔ A = z ∨ D ∩ ⦋ z / x⦌ C = ∅
10 9 ralbidv ⊢ y = A → ∀ z ∈ A B y = z ∨ ⦋ y / x⦌ C ∩ ⦋ z / x⦌ C = ∅ ↔ ∀ z ∈ A B A = z ∨ D ∩ ⦋ z / x⦌ C = ∅
11 eqeq1 ⊢ y = B → y = z ↔ B = z
12 nfcv ⊢ Ⅎ _ x B
13 nfcv ⊢ Ⅎ _ x E
14 12 13 2 csbhypf ⊢ y = B → ⦋ y / x⦌ C = E
15 14 ineq1d ⊢ y = B → ⦋ y / x⦌ C ∩ ⦋ z / x⦌ C = E ∩ ⦋ z / x⦌ C
16 15 eqeq1d ⊢ y = B → ⦋ y / x⦌ C ∩ ⦋ z / x⦌ C = ∅ ↔ E ∩ ⦋ z / x⦌ C = ∅
17 11 16 orbi12d ⊢ y = B → y = z ∨ ⦋ y / x⦌ C ∩ ⦋ z / x⦌ C = ∅ ↔ B = z ∨ E ∩ ⦋ z / x⦌ C = ∅
18 17 ralbidv ⊢ y = B → ∀ z ∈ A B y = z ∨ ⦋ y / x⦌ C ∩ ⦋ z / x⦌ C = ∅ ↔ ∀ z ∈ A B B = z ∨ E ∩ ⦋ z / x⦌ C = ∅
19 10 18 ralprg ⊢ A ∈ V ∧ B ∈ V → ∀ y ∈ A B ∀ z ∈ A B y = z ∨ ⦋ y / x⦌ C ∩ ⦋ z / x⦌ C = ∅ ↔ ∀ z ∈ A B A = z ∨ D ∩ ⦋ z / x⦌ C = ∅ ∧ ∀ z ∈ A B B = z ∨ E ∩ ⦋ z / x⦌ C = ∅
20 19 3adant3 ⊢ A ∈ V ∧ B ∈ V ∧ A ≠ B → ∀ y ∈ A B ∀ z ∈ A B y = z ∨ ⦋ y / x⦌ C ∩ ⦋ z / x⦌ C = ∅ ↔ ∀ z ∈ A B A = z ∨ D ∩ ⦋ z / x⦌ C = ∅ ∧ ∀ z ∈ A B B = z ∨ E ∩ ⦋ z / x⦌ C = ∅
21 id ⊢ z = A → z = A
22 21 eqcomd ⊢ z = A → A = z
23 22 orcd ⊢ z = A → A = z ∨ D ∩ ⦋ z / x⦌ C = ∅
24 trud ⊢ z = A → ⊤
25 23 24 2thd ⊢ z = A → A = z ∨ D ∩ ⦋ z / x⦌ C = ∅ ↔ ⊤
26 eqeq2 ⊢ z = B → A = z ↔ A = B
27 12 13 2 csbhypf ⊢ z = B → ⦋ z / x⦌ C = E
28 27 ineq2d ⊢ z = B → D ∩ ⦋ z / x⦌ C = D ∩ E
29 28 eqeq1d ⊢ z = B → D ∩ ⦋ z / x⦌ C = ∅ ↔ D ∩ E = ∅
30 26 29 orbi12d ⊢ z = B → A = z ∨ D ∩ ⦋ z / x⦌ C = ∅ ↔ A = B ∨ D ∩ E = ∅
31 25 30 ralprg ⊢ A ∈ V ∧ B ∈ V → ∀ z ∈ A B A = z ∨ D ∩ ⦋ z / x⦌ C = ∅ ↔ ⊤ ∧ A = B ∨ D ∩ E = ∅
32 31 3adant3 ⊢ A ∈ V ∧ B ∈ V ∧ A ≠ B → ∀ z ∈ A B A = z ∨ D ∩ ⦋ z / x⦌ C = ∅ ↔ ⊤ ∧ A = B ∨ D ∩ E = ∅
33 simp3 ⊢ A ∈ V ∧ B ∈ V ∧ A ≠ B → A ≠ B
34 33 neneqd ⊢ A ∈ V ∧ B ∈ V ∧ A ≠ B → ¬ A = B
35 biorf ⊢ ¬ A = B → D ∩ E = ∅ ↔ A = B ∨ D ∩ E = ∅
36 34 35 syl ⊢ A ∈ V ∧ B ∈ V ∧ A ≠ B → D ∩ E = ∅ ↔ A = B ∨ D ∩ E = ∅
37 tru ⊢ ⊤
38 37 biantrur ⊢ A = B ∨ D ∩ E = ∅ ↔ ⊤ ∧ A = B ∨ D ∩ E = ∅
39 36 38 bitrdi ⊢ A ∈ V ∧ B ∈ V ∧ A ≠ B → D ∩ E = ∅ ↔ ⊤ ∧ A = B ∨ D ∩ E = ∅
40 32 39 bitr4d ⊢ A ∈ V ∧ B ∈ V ∧ A ≠ B → ∀ z ∈ A B A = z ∨ D ∩ ⦋ z / x⦌ C = ∅ ↔ D ∩ E = ∅
41 eqeq2 ⊢ z = A → B = z ↔ B = A
42 eqcom ⊢ B = A ↔ A = B
43 41 42 bitrdi ⊢ z = A → B = z ↔ A = B
44 4 5 1 csbhypf ⊢ z = A → ⦋ z / x⦌ C = D
45 44 ineq2d ⊢ z = A → E ∩ ⦋ z / x⦌ C = E ∩ D
46 incom ⊢ E ∩ D = D ∩ E
47 45 46 eqtrdi ⊢ z = A → E ∩ ⦋ z / x⦌ C = D ∩ E
48 47 eqeq1d ⊢ z = A → E ∩ ⦋ z / x⦌ C = ∅ ↔ D ∩ E = ∅
49 43 48 orbi12d ⊢ z = A → B = z ∨ E ∩ ⦋ z / x⦌ C = ∅ ↔ A = B ∨ D ∩ E = ∅
50 id ⊢ z = B → z = B
51 50 eqcomd ⊢ z = B → B = z
52 51 orcd ⊢ z = B → B = z ∨ E ∩ ⦋ z / x⦌ C = ∅
53 trud ⊢ z = B → ⊤
54 52 53 2thd ⊢ z = B → B = z ∨ E ∩ ⦋ z / x⦌ C = ∅ ↔ ⊤
55 49 54 ralprg ⊢ A ∈ V ∧ B ∈ V → ∀ z ∈ A B B = z ∨ E ∩ ⦋ z / x⦌ C = ∅ ↔ A = B ∨ D ∩ E = ∅ ∧ ⊤
56 55 3adant3 ⊢ A ∈ V ∧ B ∈ V ∧ A ≠ B → ∀ z ∈ A B B = z ∨ E ∩ ⦋ z / x⦌ C = ∅ ↔ A = B ∨ D ∩ E = ∅ ∧ ⊤
57 37 biantru ⊢ A = B ∨ D ∩ E = ∅ ↔ A = B ∨ D ∩ E = ∅ ∧ ⊤
58 36 57 bitrdi ⊢ A ∈ V ∧ B ∈ V ∧ A ≠ B → D ∩ E = ∅ ↔ A = B ∨ D ∩ E = ∅ ∧ ⊤
59 56 58 bitr4d ⊢ A ∈ V ∧ B ∈ V ∧ A ≠ B → ∀ z ∈ A B B = z ∨ E ∩ ⦋ z / x⦌ C = ∅ ↔ D ∩ E = ∅
60 40 59 anbi12d ⊢ A ∈ V ∧ B ∈ V ∧ A ≠ B → ∀ z ∈ A B A = z ∨ D ∩ ⦋ z / x⦌ C = ∅ ∧ ∀ z ∈ A B B = z ∨ E ∩ ⦋ z / x⦌ C = ∅ ↔ D ∩ E = ∅ ∧ D ∩ E = ∅
61 20 60 bitrd ⊢ A ∈ V ∧ B ∈ V ∧ A ≠ B → ∀ y ∈ A B ∀ z ∈ A B y = z ∨ ⦋ y / x⦌ C ∩ ⦋ z / x⦌ C = ∅ ↔ D ∩ E = ∅ ∧ D ∩ E = ∅
62 disjors ⊢ Disj x ∈ A B C ↔ ∀ y ∈ A B ∀ z ∈ A B y = z ∨ ⦋ y / x⦌ C ∩ ⦋ z / x⦌ C = ∅
63 pm4.24 ⊢ D ∩ E = ∅ ↔ D ∩ E = ∅ ∧ D ∩ E = ∅
64 61 62 63 3bitr4g ⊢ A ∈ V ∧ B ∈ V ∧ A ≠ B → Disj x ∈ A B C ↔ D ∩ E = ∅