Metamath Proof Explorer


Theorem disjxrn

Description: Two ways of saying that a range Cartesian product is disjoint. (Contributed by Peter Mazsa, 17-Jun-2020) (Revised by Peter Mazsa, 21-Sep-2021)

Ref Expression
Assertion disjxrn ⊢ Disj R ⋉ S ↔ ≀ R -1 ∩ ≀ S -1 ⊆ I

Proof

Step Hyp Ref Expression
1 xrnrel ⊢ Rel ⁡ R ⋉ S
2 dfdisjALTV2 ⊢ Disj R ⋉ S ↔ ≀ R ⋉ S -1 ⊆ I ∧ Rel ⁡ R ⋉ S
3 1 2 mpbiran2 ⊢ Disj R ⋉ S ↔ ≀ R ⋉ S -1 ⊆ I
4 1cosscnvxrn ⊢ ≀ R ⋉ S -1 = ≀ R -1 ∩ ≀ S -1
5 4 sseq1i ⊢ ≀ R ⋉ S -1 ⊆ I ↔ ≀ R -1 ∩ ≀ S -1 ⊆ I
6 3 5 bitri ⊢ Disj R ⋉ S ↔ ≀ R -1 ∩ ≀ S -1 ⊆ I