Metamath Proof Explorer
Description: Two ways of saying a relation is transitive. (Contributed by RP, 22-Mar-2020)
|
|
Ref |
Expression |
|
Hypotheses |
cotr3.a |
⊢ 𝐴 = dom 𝑅 |
|
|
cotr3.b |
⊢ 𝐵 = ( 𝐴 ∩ 𝐶 ) |
|
|
cotr3.c |
⊢ 𝐶 = ran 𝑅 |
|
Assertion |
cotr3 |
⊢ ( ( 𝑅 ∘ 𝑅 ) ⊆ 𝑅 ↔ ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐵 ∀ 𝑧 ∈ 𝐶 ( ( 𝑥 𝑅 𝑦 ∧ 𝑦 𝑅 𝑧 ) → 𝑥 𝑅 𝑧 ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
cotr3.a |
⊢ 𝐴 = dom 𝑅 |
2 |
|
cotr3.b |
⊢ 𝐵 = ( 𝐴 ∩ 𝐶 ) |
3 |
|
cotr3.c |
⊢ 𝐶 = ran 𝑅 |
4 |
1
|
eqimss2i |
⊢ dom 𝑅 ⊆ 𝐴 |
5 |
1 3
|
ineq12i |
⊢ ( 𝐴 ∩ 𝐶 ) = ( dom 𝑅 ∩ ran 𝑅 ) |
6 |
2 5
|
eqtri |
⊢ 𝐵 = ( dom 𝑅 ∩ ran 𝑅 ) |
7 |
6
|
eqimss2i |
⊢ ( dom 𝑅 ∩ ran 𝑅 ) ⊆ 𝐵 |
8 |
3
|
eqimss2i |
⊢ ran 𝑅 ⊆ 𝐶 |
9 |
4 7 8
|
cotr2 |
⊢ ( ( 𝑅 ∘ 𝑅 ) ⊆ 𝑅 ↔ ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐵 ∀ 𝑧 ∈ 𝐶 ( ( 𝑥 𝑅 𝑦 ∧ 𝑦 𝑅 𝑧 ) → 𝑥 𝑅 𝑧 ) ) |