Metamath Proof Explorer


Theorem rntrcl

Description: The range of the transitive closure is equal to the range of its base relation. (Contributed by RP, 1-Nov-2020)

Ref Expression
Assertion rntrcl ( 𝑋𝑉 → ran { 𝑥 ∣ ( 𝑋𝑥 ∧ ( 𝑥𝑥 ) ⊆ 𝑥 ) } = ran 𝑋 )

Proof

Step Hyp Ref Expression
1 trclubg ( 𝑋𝑉 { 𝑥 ∣ ( 𝑋𝑥 ∧ ( 𝑥𝑥 ) ⊆ 𝑥 ) } ⊆ ( 𝑋 ∪ ( dom 𝑋 × ran 𝑋 ) ) )
2 rnss ( { 𝑥 ∣ ( 𝑋𝑥 ∧ ( 𝑥𝑥 ) ⊆ 𝑥 ) } ⊆ ( 𝑋 ∪ ( dom 𝑋 × ran 𝑋 ) ) → ran { 𝑥 ∣ ( 𝑋𝑥 ∧ ( 𝑥𝑥 ) ⊆ 𝑥 ) } ⊆ ran ( 𝑋 ∪ ( dom 𝑋 × ran 𝑋 ) ) )
3 1 2 syl ( 𝑋𝑉 → ran { 𝑥 ∣ ( 𝑋𝑥 ∧ ( 𝑥𝑥 ) ⊆ 𝑥 ) } ⊆ ran ( 𝑋 ∪ ( dom 𝑋 × ran 𝑋 ) ) )
4 rnun ran ( 𝑋 ∪ ( dom 𝑋 × ran 𝑋 ) ) = ( ran 𝑋 ∪ ran ( dom 𝑋 × ran 𝑋 ) )
5 rnxpss ran ( dom 𝑋 × ran 𝑋 ) ⊆ ran 𝑋
6 ssequn2 ( ran ( dom 𝑋 × ran 𝑋 ) ⊆ ran 𝑋 ↔ ( ran 𝑋 ∪ ran ( dom 𝑋 × ran 𝑋 ) ) = ran 𝑋 )
7 5 6 mpbi ( ran 𝑋 ∪ ran ( dom 𝑋 × ran 𝑋 ) ) = ran 𝑋
8 4 7 eqtri ran ( 𝑋 ∪ ( dom 𝑋 × ran 𝑋 ) ) = ran 𝑋
9 3 8 sseqtrdi ( 𝑋𝑉 → ran { 𝑥 ∣ ( 𝑋𝑥 ∧ ( 𝑥𝑥 ) ⊆ 𝑥 ) } ⊆ ran 𝑋 )
10 ssmin 𝑋 { 𝑥 ∣ ( 𝑋𝑥 ∧ ( 𝑥𝑥 ) ⊆ 𝑥 ) }
11 rnss ( 𝑋 { 𝑥 ∣ ( 𝑋𝑥 ∧ ( 𝑥𝑥 ) ⊆ 𝑥 ) } → ran 𝑋 ⊆ ran { 𝑥 ∣ ( 𝑋𝑥 ∧ ( 𝑥𝑥 ) ⊆ 𝑥 ) } )
12 10 11 mp1i ( 𝑋𝑉 → ran 𝑋 ⊆ ran { 𝑥 ∣ ( 𝑋𝑥 ∧ ( 𝑥𝑥 ) ⊆ 𝑥 ) } )
13 9 12 eqssd ( 𝑋𝑉 → ran { 𝑥 ∣ ( 𝑋𝑥 ∧ ( 𝑥𝑥 ) ⊆ 𝑥 ) } = ran 𝑋 )