Metamath Proof Explorer


Theorem disjif2

Description: Property of a disjoint collection: if B ( x ) and B ( Y ) = D have a common element Z , then x = Y . (Contributed by Thierry Arnoux, 6-Apr-2017)

Ref Expression
Hypotheses disjif2.1 ⊢ Ⅎ _ x A
disjif2.2 ⊢ Ⅎ _ x C
disjif2.3 ⊢ x = Y → B = C
Assertion disjif2 ⊢ Disj x ∈ A B ∧ x ∈ A ∧ Y ∈ A ∧ Z ∈ B ∧ Z ∈ C → x = Y

Proof

Step Hyp Ref Expression
1 disjif2.1 ⊢ Ⅎ _ x A
2 disjif2.2 ⊢ Ⅎ _ x C
3 disjif2.3 ⊢ x = Y → B = C
4 inelcm ⊢ Z ∈ B ∧ Z ∈ C → B ∩ C ≠ ∅
5 1 disjorsf ⊢ Disj x ∈ A B ↔ ∀ y ∈ A ∀ z ∈ A y = z ∨ ⦋ y / x⦌ B ∩ ⦋ z / x⦌ B = ∅
6 equequ1 ⊢ y = x → y = z ↔ x = z
7 csbeq1 ⊢ y = x → ⦋ y / x⦌ B = ⦋ x / x⦌ B
8 csbid ⊢ ⦋ x / x⦌ B = B
9 7 8 eqtrdi ⊢ y = x → ⦋ y / x⦌ B = B
10 9 ineq1d ⊢ y = x → ⦋ y / x⦌ B ∩ ⦋ z / x⦌ B = B ∩ ⦋ z / x⦌ B
11 10 eqeq1d ⊢ y = x → ⦋ y / x⦌ B ∩ ⦋ z / x⦌ B = ∅ ↔ B ∩ ⦋ z / x⦌ B = ∅
12 6 11 orbi12d ⊢ y = x → y = z ∨ ⦋ y / x⦌ B ∩ ⦋ z / x⦌ B = ∅ ↔ x = z ∨ B ∩ ⦋ z / x⦌ B = ∅
13 eqeq2 ⊢ z = Y → x = z ↔ x = Y
14 nfcv ⊢ Ⅎ _ x Y
15 14 2 3 csbhypf ⊢ z = Y → ⦋ z / x⦌ B = C
16 15 ineq2d ⊢ z = Y → B ∩ ⦋ z / x⦌ B = B ∩ C
17 16 eqeq1d ⊢ z = Y → B ∩ ⦋ z / x⦌ B = ∅ ↔ B ∩ C = ∅
18 13 17 orbi12d ⊢ z = Y → x = z ∨ B ∩ ⦋ z / x⦌ B = ∅ ↔ x = Y ∨ B ∩ C = ∅
19 12 18 rspc2v ⊢ x ∈ A ∧ Y ∈ A → ∀ y ∈ A ∀ z ∈ A y = z ∨ ⦋ y / x⦌ B ∩ ⦋ z / x⦌ B = ∅ → x = Y ∨ B ∩ C = ∅
20 5 19 biimtrid ⊢ x ∈ A ∧ Y ∈ A → Disj x ∈ A B → x = Y ∨ B ∩ C = ∅
21 20 impcom ⊢ Disj x ∈ A B ∧ x ∈ A ∧ Y ∈ A → x = Y ∨ B ∩ C = ∅
22 21 ord ⊢ Disj x ∈ A B ∧ x ∈ A ∧ Y ∈ A → ¬ x = Y → B ∩ C = ∅
23 22 necon1ad ⊢ Disj x ∈ A B ∧ x ∈ A ∧ Y ∈ A → B ∩ C ≠ ∅ → x = Y
24 23 3impia ⊢ Disj x ∈ A B ∧ x ∈ A ∧ Y ∈ A ∧ B ∩ C ≠ ∅ → x = Y
25 4 24 syl3an3 ⊢ Disj x ∈ A B ∧ x ∈ A ∧ Y ∈ A ∧ Z ∈ B ∧ Z ∈ C → x = Y