Metamath Proof Explorer


Theorem tcss

Description: The transitive closure function inherits the subset relation. (Contributed by Mario Carneiro, 23-Jun-2013)

Ref Expression
Hypothesis tc2.1 𝐴 ∈ V
Assertion tcss ( 𝐵𝐴 → ( TC ‘ 𝐵 ) ⊆ ( TC ‘ 𝐴 ) )

Proof

Step Hyp Ref Expression
1 tc2.1 𝐴 ∈ V
2 1 ssex ( 𝐵𝐴𝐵 ∈ V )
3 tcvalg ( 𝐵 ∈ V → ( TC ‘ 𝐵 ) = { 𝑥 ∣ ( 𝐵𝑥 ∧ Tr 𝑥 ) } )
4 2 3 syl ( 𝐵𝐴 → ( TC ‘ 𝐵 ) = { 𝑥 ∣ ( 𝐵𝑥 ∧ Tr 𝑥 ) } )
5 sstr2 ( 𝐵𝐴 → ( 𝐴𝑥𝐵𝑥 ) )
6 5 anim1d ( 𝐵𝐴 → ( ( 𝐴𝑥 ∧ Tr 𝑥 ) → ( 𝐵𝑥 ∧ Tr 𝑥 ) ) )
7 6 ss2abdv ( 𝐵𝐴 → { 𝑥 ∣ ( 𝐴𝑥 ∧ Tr 𝑥 ) } ⊆ { 𝑥 ∣ ( 𝐵𝑥 ∧ Tr 𝑥 ) } )
8 intss ( { 𝑥 ∣ ( 𝐴𝑥 ∧ Tr 𝑥 ) } ⊆ { 𝑥 ∣ ( 𝐵𝑥 ∧ Tr 𝑥 ) } → { 𝑥 ∣ ( 𝐵𝑥 ∧ Tr 𝑥 ) } ⊆ { 𝑥 ∣ ( 𝐴𝑥 ∧ Tr 𝑥 ) } )
9 7 8 syl ( 𝐵𝐴 { 𝑥 ∣ ( 𝐵𝑥 ∧ Tr 𝑥 ) } ⊆ { 𝑥 ∣ ( 𝐴𝑥 ∧ Tr 𝑥 ) } )
10 tcvalg ( 𝐴 ∈ V → ( TC ‘ 𝐴 ) = { 𝑥 ∣ ( 𝐴𝑥 ∧ Tr 𝑥 ) } )
11 1 10 ax-mp ( TC ‘ 𝐴 ) = { 𝑥 ∣ ( 𝐴𝑥 ∧ Tr 𝑥 ) }
12 9 11 sseqtrrdi ( 𝐵𝐴 { 𝑥 ∣ ( 𝐵𝑥 ∧ Tr 𝑥 ) } ⊆ ( TC ‘ 𝐴 ) )
13 4 12 eqsstrd ( 𝐵𝐴 → ( TC ‘ 𝐵 ) ⊆ ( TC ‘ 𝐴 ) )