Metamath Proof Explorer


Theorem sspreima

Description: The preimage of a subset is a subset of the preimage. (Contributed by Brendan Leahy, 23-Sep-2017)

Ref Expression
Assertion sspreima ⊢ Fun ⁡ F ∧ A ⊆ B → F -1 A ⊆ F -1 B

Proof

Step Hyp Ref Expression
1 inpreima ⊢ Fun ⁡ F → F -1 A ∩ B = F -1 A ∩ F -1 B
2 dfss2 ⊢ A ⊆ B ↔ A ∩ B = A
3 2 biimpi ⊢ A ⊆ B → A ∩ B = A
4 3 imaeq2d ⊢ A ⊆ B → F -1 A ∩ B = F -1 A
5 1 4 sylan9req ⊢ Fun ⁡ F ∧ A ⊆ B → F -1 A ∩ F -1 B = F -1 A
6 dfss2 ⊢ F -1 A ⊆ F -1 B ↔ F -1 A ∩ F -1 B = F -1 A
7 5 6 sylibr ⊢ Fun ⁡ F ∧ A ⊆ B → F -1 A ⊆ F -1 B