Metamath Proof Explorer

Theorem inxpss3

Description: Two ways to say that an intersection with a Cartesian product is a subclass (see also inxpss ). (Contributed by Peter Mazsa, 8-Mar-2019)

		Ref	Expression
	Assertion	inxpss3	$⊢ \forall x \forall y (x (R \cap (A \times B)) y \to x (S \cap (A \times B)) y) \leftrightarrow \forall x \in A \forall y \in B (x R y \to x S y)$

Proof

Step	Hyp	Ref	Expression
1		brinxp2	$⊢ x (R \cap (A \times B)) y \leftrightarrow ((x \in A \land y \in B) \land x R y)$
2		brinxp2	$⊢ x (S \cap (A \times B)) y \leftrightarrow ((x \in A \land y \in B) \land x S y)$
3	1 2	imbi12i	$⊢ (x (R \cap (A \times B)) y \to x (S \cap (A \times B)) y) \leftrightarrow (((x \in A \land y \in B) \land x R y) \to ((x \in A \land y \in B) \land x S y))$
4		imdistan	$⊢ ((x \in A \land y \in B) \to (x R y \to x S y)) \leftrightarrow (((x \in A \land y \in B) \land x R y) \to ((x \in A \land y \in B) \land x S y))$
5	3 4	bitr4i	$⊢ (x (R \cap (A \times B)) y \to x (S \cap (A \times B)) y) \leftrightarrow ((x \in A \land y \in B) \to (x R y \to x S y))$
6	5	2albii	$⊢ \forall x \forall y (x (R \cap (A \times B)) y \to x (S \cap (A \times B)) y) \leftrightarrow \forall x \forall y ((x \in A \land y \in B) \to (x R y \to x S y))$
7		r2al	$⊢ \forall x \in A \forall y \in B (x R y \to x S y) \leftrightarrow \forall x \forall y ((x \in A \land y \in B) \to (x R y \to x S y))$
8	6 7	bitr4i	$⊢ \forall x \forall y (x (R \cap (A \times B)) y \to x (S \cap (A \times B)) y) \leftrightarrow \forall x \in A \forall y \in B (x R y \to x S y)$