Metamath Proof Explorer


Theorem bj-xpnzex

Description: If the first factor of a product is nonempty, and the product is a set, then the second factor is a set. UPDATE: this is actually the curried (exported) form of xpexcnv (up to commutation in the product). (Contributed by BJ, 6-Oct-2018) (Proof modification is discouraged.)

Ref Expression
Assertion bj-xpnzex ⊢ A ≠ ∅ → A × B ∈ V → B ∈ V

Proof

Step Hyp Ref Expression
1 0ex ⊢ ∅ ∈ V
2 eleq1a ⊢ ∅ ∈ V → B = ∅ → B ∈ V
3 1 2 ax-mp ⊢ B = ∅ → B ∈ V
4 3 a1d ⊢ B = ∅ → A × B ∈ V → B ∈ V
5 4 a1d ⊢ B = ∅ → A ≠ ∅ → A × B ∈ V → B ∈ V
6 xpnz ⊢ A ≠ ∅ ∧ B ≠ ∅ ↔ A × B ≠ ∅
7 xpexr2 ⊢ A × B ∈ V ∧ A × B ≠ ∅ → A ∈ V ∧ B ∈ V
8 7 simprd ⊢ A × B ∈ V ∧ A × B ≠ ∅ → B ∈ V
9 8 expcom ⊢ A × B ≠ ∅ → A × B ∈ V → B ∈ V
10 6 9 sylbi ⊢ A ≠ ∅ ∧ B ≠ ∅ → A × B ∈ V → B ∈ V
11 10 expcom ⊢ B ≠ ∅ → A ≠ ∅ → A × B ∈ V → B ∈ V
12 5 11 pm2.61ine ⊢ A ≠ ∅ → A × B ∈ V → B ∈ V