Metamath Proof Explorer


Theorem predrelss

Description: Subset carries from relation to predecessor class. (Contributed by Scott Fenton, 25-Nov-2024)

Ref Expression
Assertion predrelss ⊢ R ⊆ S → Pred R A X ⊆ Pred S A X

Proof

Step Hyp Ref Expression
1 cnvss ⊢ R ⊆ S → R -1 ⊆ S -1
2 imass1 ⊢ R -1 ⊆ S -1 → R -1 X ⊆ S -1 X
3 sslin ⊢ R -1 X ⊆ S -1 X → A ∩ R -1 X ⊆ A ∩ S -1 X
4 1 2 3 3syl ⊢ R ⊆ S → A ∩ R -1 X ⊆ A ∩ S -1 X
5 df-pred ⊢ Pred R A X = A ∩ R -1 X
6 df-pred ⊢ Pred S A X = A ∩ S -1 X
7 4 5 6 3sstr4g ⊢ R ⊆ S → Pred R A X ⊆ Pred S A X