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 ⊆ ◡ 𝑆 )
2 imass1 ⊢ ( ◡ 𝑅 ⊆ ◡ 𝑆 → ( ◡ 𝑅 “ { 𝑋 } ) ⊆ ( ◡ 𝑆 “ { 𝑋 } ) )
3 sslin ⊢ ( ( ◡ 𝑅 “ { 𝑋 } ) ⊆ ( ◡ 𝑆 “ { 𝑋 } ) → ( 𝐴 ∩ ( ◡ 𝑅 “ { 𝑋 } ) ) ⊆ ( 𝐴 ∩ ( ◡ 𝑆 “ { 𝑋 } ) ) )
4 1 2 3 3syl ⊢ ( 𝑅 ⊆ 𝑆 → ( 𝐴 ∩ ( ◡ 𝑅 “ { 𝑋 } ) ) ⊆ ( 𝐴 ∩ ( ◡ 𝑆 “ { 𝑋 } ) ) )
5 df-pred ⊢ Pred ( 𝑅 , 𝐴 , 𝑋 ) = ( 𝐴 ∩ ( ◡ 𝑅 “ { 𝑋 } ) )
6 df-pred ⊢ Pred ( 𝑆 , 𝐴 , 𝑋 ) = ( 𝐴 ∩ ( ◡ 𝑆 “ { 𝑋 } ) )
7 4 5 6 3sstr4g ⊢ ( 𝑅 ⊆ 𝑆 → Pred ( 𝑅 , 𝐴 , 𝑋 ) ⊆ Pred ( 𝑆 , 𝐴 , 𝑋 ) )