Metamath Proof Explorer


Theorem tskxpss

Description: A Cartesian product of two parts of a Tarski class is a part of the class. (Contributed by FL, 22-Feb-2011) (Proof shortened by Mario Carneiro, 20-Jun-2013)

Ref Expression
Assertion tskxpss ⊢ T ∈ Tarski ∧ A ⊆ T ∧ B ⊆ T → A × B ⊆ T

Proof

Step Hyp Ref Expression
1 elxp2 ⊢ z ∈ T × T ↔ ∃ x ∈ T ∃ y ∈ T z = x y
2 tskop ⊢ T ∈ Tarski ∧ x ∈ T ∧ y ∈ T → x y ∈ T
3 eleq1a ⊢ x y ∈ T → z = x y → z ∈ T
4 2 3 syl ⊢ T ∈ Tarski ∧ x ∈ T ∧ y ∈ T → z = x y → z ∈ T
5 4 3expib ⊢ T ∈ Tarski → x ∈ T ∧ y ∈ T → z = x y → z ∈ T
6 5 rexlimdvv ⊢ T ∈ Tarski → ∃ x ∈ T ∃ y ∈ T z = x y → z ∈ T
7 1 6 biimtrid ⊢ T ∈ Tarski → z ∈ T × T → z ∈ T
8 7 ssrdv ⊢ T ∈ Tarski → T × T ⊆ T
9 xpss12 ⊢ A ⊆ T ∧ B ⊆ T → A × B ⊆ T × T
10 sstr ⊢ A × B ⊆ T × T ∧ T × T ⊆ T → A × B ⊆ T
11 10 expcom ⊢ T × T ⊆ T → A × B ⊆ T × T → A × B ⊆ T
12 8 9 11 syl2im ⊢ T ∈ Tarski → A ⊆ T ∧ B ⊆ T → A × B ⊆ T
13 12 3impib ⊢ T ∈ Tarski ∧ A ⊆ T ∧ B ⊆ T → A × B ⊆ T