Metamath Proof Explorer


Theorem trclfvss

Description: The transitive closure (as a relation) of a subclass is a subclass of the transitive closure. (Contributed by RP, 3-May-2020)

Ref Expression
Assertion trclfvss ( ( 𝑅 ∈ 𝑉 ∧ 𝑆 ∈ 𝑊 ∧ 𝑅 ⊆ 𝑆 ) → ( t+ ‘ 𝑅 ) ⊆ ( t+ ‘ 𝑆 ) )

Proof

Step Hyp Ref Expression
1 trclsslem ⊢ ( 𝑅 ⊆ 𝑆 → ∩ { 𝑟 ∣ ( 𝑅 ⊆ 𝑟 ∧ ( 𝑟 ∘ 𝑟 ) ⊆ 𝑟 ) } ⊆ ∩ { 𝑟 ∣ ( 𝑆 ⊆ 𝑟 ∧ ( 𝑟 ∘ 𝑟 ) ⊆ 𝑟 ) } )
2 1 3ad2ant3 ⊢ ( ( 𝑅 ∈ 𝑉 ∧ 𝑆 ∈ 𝑊 ∧ 𝑅 ⊆ 𝑆 ) → ∩ { 𝑟 ∣ ( 𝑅 ⊆ 𝑟 ∧ ( 𝑟 ∘ 𝑟 ) ⊆ 𝑟 ) } ⊆ ∩ { 𝑟 ∣ ( 𝑆 ⊆ 𝑟 ∧ ( 𝑟 ∘ 𝑟 ) ⊆ 𝑟 ) } )
3 trclfv ⊢ ( 𝑅 ∈ 𝑉 → ( t+ ‘ 𝑅 ) = ∩ { 𝑟 ∣ ( 𝑅 ⊆ 𝑟 ∧ ( 𝑟 ∘ 𝑟 ) ⊆ 𝑟 ) } )
4 3 3ad2ant1 ⊢ ( ( 𝑅 ∈ 𝑉 ∧ 𝑆 ∈ 𝑊 ∧ 𝑅 ⊆ 𝑆 ) → ( t+ ‘ 𝑅 ) = ∩ { 𝑟 ∣ ( 𝑅 ⊆ 𝑟 ∧ ( 𝑟 ∘ 𝑟 ) ⊆ 𝑟 ) } )
5 trclfv ⊢ ( 𝑆 ∈ 𝑊 → ( t+ ‘ 𝑆 ) = ∩ { 𝑟 ∣ ( 𝑆 ⊆ 𝑟 ∧ ( 𝑟 ∘ 𝑟 ) ⊆ 𝑟 ) } )
6 5 3ad2ant2 ⊢ ( ( 𝑅 ∈ 𝑉 ∧ 𝑆 ∈ 𝑊 ∧ 𝑅 ⊆ 𝑆 ) → ( t+ ‘ 𝑆 ) = ∩ { 𝑟 ∣ ( 𝑆 ⊆ 𝑟 ∧ ( 𝑟 ∘ 𝑟 ) ⊆ 𝑟 ) } )
7 2 4 6 3sstr4d ⊢ ( ( 𝑅 ∈ 𝑉 ∧ 𝑆 ∈ 𝑊 ∧ 𝑅 ⊆ 𝑆 ) → ( t+ ‘ 𝑅 ) ⊆ ( t+ ‘ 𝑆 ) )