Metamath Proof Explorer


Theorem inxpssidinxp

Description: Two ways to say that intersections with Cartesian products are in a subclass relation, special case of inxpss2 . (Contributed by Peter Mazsa, 4-Jul-2019)

Ref Expression
Assertion inxpssidinxp ⊢ R ∩ A × B ⊆ I ∩ A × B ↔ ∀ x ∈ A ∀ y ∈ B x R y → x = y

Proof

Step Hyp Ref Expression
1 inxpss2 ⊢ R ∩ A × B ⊆ I ∩ A × B ↔ ∀ x ∈ A ∀ y ∈ B x R y → x I y
2 ideqg ⊢ y ∈ V → x I y ↔ x = y
3 2 elv ⊢ x I y ↔ x = y
4 3 imbi2i ⊢ x R y → x I y ↔ x R y → x = y
5 4 2ralbii ⊢ ∀ x ∈ A ∀ y ∈ B x R y → x I y ↔ ∀ x ∈ A ∀ y ∈ B x R y → x = y
6 1 5 bitri ⊢ R ∩ A × B ⊆ I ∩ A × B ↔ ∀ x ∈ A ∀ y ∈ B x R y → x = y