Metamath Proof Explorer


Theorem ssxpb

Description: A Cartesian product subclass relationship is equivalent to the conjunction of the analogous relationships for the factors. (Contributed by NM, 17-Dec-2008)

Ref Expression
Assertion ssxpb ⊢ A × B ≠ ∅ → A × B ⊆ C × D ↔ A ⊆ C ∧ B ⊆ D

Proof

Step Hyp Ref Expression
1 xpnz ⊢ A ≠ ∅ ∧ B ≠ ∅ ↔ A × B ≠ ∅
2 dmxp ⊢ B ≠ ∅ → dom ⁡ A × B = A
3 2 adantl ⊢ A ≠ ∅ ∧ B ≠ ∅ → dom ⁡ A × B = A
4 1 3 sylbir ⊢ A × B ≠ ∅ → dom ⁡ A × B = A
5 4 adantr ⊢ A × B ≠ ∅ ∧ A × B ⊆ C × D → dom ⁡ A × B = A
6 dmss ⊢ A × B ⊆ C × D → dom ⁡ A × B ⊆ dom ⁡ C × D
7 6 adantl ⊢ A × B ≠ ∅ ∧ A × B ⊆ C × D → dom ⁡ A × B ⊆ dom ⁡ C × D
8 5 7 eqsstrrd ⊢ A × B ≠ ∅ ∧ A × B ⊆ C × D → A ⊆ dom ⁡ C × D
9 dmxpss ⊢ dom ⁡ C × D ⊆ C
10 8 9 sstrdi ⊢ A × B ≠ ∅ ∧ A × B ⊆ C × D → A ⊆ C
11 rnxp ⊢ A ≠ ∅ → ran ⁡ A × B = B
12 11 adantr ⊢ A ≠ ∅ ∧ B ≠ ∅ → ran ⁡ A × B = B
13 1 12 sylbir ⊢ A × B ≠ ∅ → ran ⁡ A × B = B
14 13 adantr ⊢ A × B ≠ ∅ ∧ A × B ⊆ C × D → ran ⁡ A × B = B
15 rnss ⊢ A × B ⊆ C × D → ran ⁡ A × B ⊆ ran ⁡ C × D
16 15 adantl ⊢ A × B ≠ ∅ ∧ A × B ⊆ C × D → ran ⁡ A × B ⊆ ran ⁡ C × D
17 14 16 eqsstrrd ⊢ A × B ≠ ∅ ∧ A × B ⊆ C × D → B ⊆ ran ⁡ C × D
18 rnxpss ⊢ ran ⁡ C × D ⊆ D
19 17 18 sstrdi ⊢ A × B ≠ ∅ ∧ A × B ⊆ C × D → B ⊆ D
20 10 19 jca ⊢ A × B ≠ ∅ ∧ A × B ⊆ C × D → A ⊆ C ∧ B ⊆ D
21 20 ex ⊢ A × B ≠ ∅ → A × B ⊆ C × D → A ⊆ C ∧ B ⊆ D
22 xpss12 ⊢ A ⊆ C ∧ B ⊆ D → A × B ⊆ C × D
23 21 22 impbid1 ⊢ A × B ≠ ∅ → A × B ⊆ C × D ↔ A ⊆ C ∧ B ⊆ D