Metamath Proof Explorer


Theorem altxpeq1

Description: Equality for alternate Cartesian products. (Contributed by Scott Fenton, 24-Mar-2012)

Ref Expression
Assertion altxpeq1 ⊢ A = B → A ×× C = B ×× C

Proof

Step Hyp Ref Expression
1 rexeq ⊢ A = B → ∃ x ∈ A ∃ y ∈ C z = x y ↔ ∃ x ∈ B ∃ y ∈ C z = x y
2 1 abbidv ⊢ A = B → z | ∃ x ∈ A ∃ y ∈ C z = x y = z | ∃ x ∈ B ∃ y ∈ C z = x y
3 df-altxp ⊢ A ×× C = z | ∃ x ∈ A ∃ y ∈ C z = x y
4 df-altxp ⊢ B ×× C = z | ∃ x ∈ B ∃ y ∈ C z = x y
5 2 3 4 3eqtr4g ⊢ A = B → A ×× C = B ×× C