Metamath Proof Explorer


Theorem disji2

Description: Property of a disjoint collection: if B ( X ) = C and B ( Y ) = D , and X =/= Y , then C and D are disjoint. (Contributed by Mario Carneiro, 14-Nov-2016)

Ref Expression
Hypotheses disji.1 ⊢ x = X → B = C
disji.2 ⊢ x = Y → B = D
Assertion disji2 ⊢ Disj x ∈ A B ∧ X ∈ A ∧ Y ∈ A ∧ X ≠ Y → C ∩ D = ∅

Proof

Step Hyp Ref Expression
1 disji.1 ⊢ x = X → B = C
2 disji.2 ⊢ x = Y → B = D
3 df-ne ⊢ X ≠ Y ↔ ¬ X = Y
4 disjors ⊢ Disj x ∈ A B ↔ ∀ y ∈ A ∀ z ∈ A y = z ∨ ⦋ y / x⦌ B ∩ ⦋ z / x⦌ B = ∅
5 eqeq1 ⊢ y = X → y = z ↔ X = z
6 nfcv ⊢ Ⅎ _ x X
7 nfcv ⊢ Ⅎ _ x C
8 6 7 1 csbhypf ⊢ y = X → ⦋ y / x⦌ B = C
9 8 ineq1d ⊢ y = X → ⦋ y / x⦌ B ∩ ⦋ z / x⦌ B = C ∩ ⦋ z / x⦌ B
10 9 eqeq1d ⊢ y = X → ⦋ y / x⦌ B ∩ ⦋ z / x⦌ B = ∅ ↔ C ∩ ⦋ z / x⦌ B = ∅
11 5 10 orbi12d ⊢ y = X → y = z ∨ ⦋ y / x⦌ B ∩ ⦋ z / x⦌ B = ∅ ↔ X = z ∨ C ∩ ⦋ z / x⦌ B = ∅
12 eqeq2 ⊢ z = Y → X = z ↔ X = Y
13 nfcv ⊢ Ⅎ _ x Y
14 nfcv ⊢ Ⅎ _ x D
15 13 14 2 csbhypf ⊢ z = Y → ⦋ z / x⦌ B = D
16 15 ineq2d ⊢ z = Y → C ∩ ⦋ z / x⦌ B = C ∩ D
17 16 eqeq1d ⊢ z = Y → C ∩ ⦋ z / x⦌ B = ∅ ↔ C ∩ D = ∅
18 12 17 orbi12d ⊢ z = Y → X = z ∨ C ∩ ⦋ z / x⦌ B = ∅ ↔ X = Y ∨ C ∩ D = ∅
19 11 18 rspc2v ⊢ X ∈ A ∧ Y ∈ A → ∀ y ∈ A ∀ z ∈ A y = z ∨ ⦋ y / x⦌ B ∩ ⦋ z / x⦌ B = ∅ → X = Y ∨ C ∩ D = ∅
20 4 19 biimtrid ⊢ X ∈ A ∧ Y ∈ A → Disj x ∈ A B → X = Y ∨ C ∩ D = ∅
21 20 impcom ⊢ Disj x ∈ A B ∧ X ∈ A ∧ Y ∈ A → X = Y ∨ C ∩ D = ∅
22 21 ord ⊢ Disj x ∈ A B ∧ X ∈ A ∧ Y ∈ A → ¬ X = Y → C ∩ D = ∅
23 3 22 biimtrid ⊢ Disj x ∈ A B ∧ X ∈ A ∧ Y ∈ A → X ≠ Y → C ∩ D = ∅
24 23 3impia ⊢ Disj x ∈ A B ∧ X ∈ A ∧ Y ∈ A ∧ X ≠ Y → C ∩ D = ∅