Theorem bj-xpnzex 34275

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 7628 (up to commutation in the product). (Contributed by BJ, 6-Oct-2018.) (Proof modification is discouraged.)

Assertion

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

Proof of Theorem bj-xpnzex

Step	Hyp	Ref	Expression
1		0ex 5214	. . . . 5 ⊢ ∅ ∈ V
2		eleq1a 2911	. . . . 5 ⊢ (∅ ∈ V → (𝐵 = ∅ → 𝐵 ∈ V))
3	1, 2	ax-mp 5	. . . 4 ⊢ (𝐵 = ∅ → 𝐵 ∈ V)
4	3	a1d 25	. . 3 ⊢ (𝐵 = ∅ → ((𝐴 × 𝐵) ∈ 𝑉 → 𝐵 ∈ V))
5	4	a1d 25	. 2 ⊢ (𝐵 = ∅ → (𝐴 ≠ ∅ → ((𝐴 × 𝐵) ∈ 𝑉 → 𝐵 ∈ V)))
6		xpnz 6019	. . . 4 ⊢ ((𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅) ↔ (𝐴 × 𝐵) ≠ ∅)
7		xpexr2 7627	. . . . . 6 ⊢ (((𝐴 × 𝐵) ∈ 𝑉 ∧ (𝐴 × 𝐵) ≠ ∅) → (𝐴 ∈ V ∧ 𝐵 ∈ V))
8	7	simprd 498	. . . . 5 ⊢ (((𝐴 × 𝐵) ∈ 𝑉 ∧ (𝐴 × 𝐵) ≠ ∅) → 𝐵 ∈ V)
9	8	expcom 416	. . . 4 ⊢ ((𝐴 × 𝐵) ≠ ∅ → ((𝐴 × 𝐵) ∈ 𝑉 → 𝐵 ∈ V))
10	6, 9	sylbi 219	. . 3 ⊢ ((𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅) → ((𝐴 × 𝐵) ∈ 𝑉 → 𝐵 ∈ V))
11	10	expcom 416	. 2 ⊢ (𝐵 ≠ ∅ → (𝐴 ≠ ∅ → ((𝐴 × 𝐵) ∈ 𝑉 → 𝐵 ∈ V)))
12	5, 11	pm2.61ine 3103	1 ⊢ (𝐴 ≠ ∅ → ((𝐴 × 𝐵) ∈ 𝑉 → 𝐵 ∈ V))

Syntax hints:   → wi 4   ∧ wa 398   = wceq 1536   ∈ wcel 2113   ≠ wne 3019  Vcvv 3497  ∅c0 4294   × cxp 5556

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1969  ax-7 2014  ax-8 2115  ax-9 2123  ax-10 2144  ax-11 2160  ax-12 2176  ax-ext 2796  ax-sep 5206  ax-nul 5213  ax-pr 5333  ax-un 7464

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-3an 1085  df-tru 1539  df-ex 1780  df-nf 1784  df-sb 2069  df-mo 2621  df-eu 2653  df-clab 2803  df-cleq 2817  df-clel 2896  df-nfc 2966  df-ne 3020  df-ral 3146  df-rex 3147  df-rab 3150  df-v 3499  df-dif 3942  df-un 3944  df-in 3946  df-ss 3955  df-nul 4295  df-if 4471  df-sn 4571  df-pr 4573  df-op 4577  df-uni 4842  df-br 5070  df-opab 5132  df-xp 5564  df-rel 5565  df-cnv 5566  df-dm 5568  df-rn 5569

This theorem is referenced by:  bj-xpnzexb  34277