Metamath Proof Explorer


Theorem predisj

Description: Preimages of disjoint sets are disjoint. (Contributed by Zhi Wang, 9-Sep-2024)

Ref Expression
Hypotheses predisj.1 ⊢ φ → Fun ⁡ F
predisj.2 ⊢ φ → A ∩ B = ∅
predisj.3 ⊢ φ → S ⊆ F -1 A
predisj.4 ⊢ φ → T ⊆ F -1 B
Assertion predisj ⊢ φ → S ∩ T = ∅

Proof

Step Hyp Ref Expression
1 predisj.1 ⊢ φ → Fun ⁡ F
2 predisj.2 ⊢ φ → A ∩ B = ∅
3 predisj.3 ⊢ φ → S ⊆ F -1 A
4 predisj.4 ⊢ φ → T ⊆ F -1 B
5 inpreima ⊢ Fun ⁡ F → F -1 A ∩ B = F -1 A ∩ F -1 B
6 1 5 syl ⊢ φ → F -1 A ∩ B = F -1 A ∩ F -1 B
7 2 imaeq2d ⊢ φ → F -1 A ∩ B = F -1 ∅
8 ima0 ⊢ F -1 ∅ = ∅
9 7 8 eqtrdi ⊢ φ → F -1 A ∩ B = ∅
10 6 9 eqtr3d ⊢ φ → F -1 A ∩ F -1 B = ∅
11 3 10 ssdisjd ⊢ φ → S ∩ F -1 B = ∅
12 4 11 ssdisjdr ⊢ φ → S ∩ T = ∅