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 =