Metamath Proof Explorer


Theorem xppss12

Description: Proper subset theorem for Cartesian product. (Contributed by Steven Nguyen, 17-Jul-2022)

Ref Expression
Assertion xppss12 ⊢ A ⊂ B ∧ C ⊂ D → A × C ⊂ B × D

Proof

Step Hyp Ref Expression
1 pssss ⊢ A ⊂ B → A ⊆ B
2 pssss ⊢ C ⊂ D → C ⊆ D
3 xpss12 ⊢ A ⊆ B ∧ C ⊆ D → A × C ⊆ B × D
4 1 2 3 syl2an ⊢ A ⊂ B ∧ C ⊂ D → A × C ⊆ B × D
5 simpl ⊢ A ⊂ B ∧ C ⊂ D → A ⊂ B
6 pssne ⊢ A ⊂ B → A ≠ B
7 6 necomd ⊢ A ⊂ B → B ≠ A
8 neneq ⊢ B ≠ A → ¬ B = A
9 8 intnanrd ⊢ B ≠ A → ¬ B = A ∧ D = C
10 5 7 9 3syl ⊢ A ⊂ B ∧ C ⊂ D → ¬ B = A ∧ D = C
11 pssn0 ⊢ A ⊂ B → B ≠ ∅
12 pssn0 ⊢ C ⊂ D → D ≠ ∅
13 xp11 ⊢ B ≠ ∅ ∧ D ≠ ∅ → B × D = A × C ↔ B = A ∧ D = C
14 11 12 13 syl2an ⊢ A ⊂ B ∧ C ⊂ D → B × D = A × C ↔ B = A ∧ D = C
15 10 14 mtbird ⊢ A ⊂ B ∧ C ⊂ D → ¬ B × D = A × C
16 neqne ⊢ ¬ B × D = A × C → B × D ≠ A × C
17 16 necomd ⊢ ¬ B × D = A × C → A × C ≠ B × D
18 15 17 syl ⊢ A ⊂ B ∧ C ⊂ D → A × C ≠ B × D
19 df-pss ⊢ A × C ⊂ B × D ↔ A × C ⊆ B × D ∧ A × C ≠ B × D
20 4 18 19 sylanbrc ⊢ A ⊂ B ∧ C ⊂ D → A × C ⊂ B × D