Theorem bj-xpnzex 34271

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 7624 (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 5210	. . . . 5 ⊢ ∅ ∈ V
2		eleq1a 2908	. . . . 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 6015	. . . 4 ⊢ ((𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅) ↔ (𝐴 × 𝐵) ≠ ∅)
7		xpexr2 7623	. . . . . 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 3100	1 ⊢ (𝐴 ≠ ∅ → ((𝐴 × 𝐵) ∈ 𝑉 → 𝐵 ∈ V))

Syntax hints:   → wi 4   ∧ wa 398   = wceq 1533   ∈ wcel 2110   ≠ wne 3016  Vcvv 3494  ∅c0 4290   × cxp 5552

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1907  ax-6 1966  ax-7 2011  ax-8 2112  ax-9 2120  ax-10 2141  ax-11 2157  ax-12 2173  ax-ext 2793  ax-sep 5202  ax-nul 5209  ax-pr 5329  ax-un 7460

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-3an 1085  df-tru 1536  df-ex 1777  df-nf 1781  df-sb 2066  df-mo 2618  df-eu 2650  df-clab 2800  df-cleq 2814  df-clel 2893  df-nfc 2963  df-ne 3017  df-ral 3143  df-rex 3144  df-rab 3147  df-v 3496  df-dif 3938  df-un 3940  df-in 3942  df-ss 3951  df-nul 4291  df-if 4467  df-sn 4567  df-pr 4569  df-op 4573  df-uni 4838  df-br 5066  df-opab 5128  df-xp 5560  df-rel 5561  df-cnv 5562  df-dm 5564  df-rn 5565

This theorem is referenced by:  bj-xpnzexb  34273