Theorem bj-xpnzex 36942

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 7943 (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 5313	. . . . 5 ⊢ ∅ ∈ V
2		eleq1a 2834	. . . . 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 6181	. . . 4 ⊢ ((𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅) ↔ (𝐴 × 𝐵) ≠ ∅)
7		xpexr2 7942	. . . . . 6 ⊢ (((𝐴 × 𝐵) ∈ 𝑉 ∧ (𝐴 × 𝐵) ≠ ∅) → (𝐴 ∈ V ∧ 𝐵 ∈ V))
8	7	simprd 495	. . . . 5 ⊢ (((𝐴 × 𝐵) ∈ 𝑉 ∧ (𝐴 × 𝐵) ≠ ∅) → 𝐵 ∈ V)
9	8	expcom 413	. . . 4 ⊢ ((𝐴 × 𝐵) ≠ ∅ → ((𝐴 × 𝐵) ∈ 𝑉 → 𝐵 ∈ V))
10	6, 9	sylbi 217	. . 3 ⊢ ((𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅) → ((𝐴 × 𝐵) ∈ 𝑉 → 𝐵 ∈ V))
11	10	expcom 413	. 2 ⊢ (𝐵 ≠ ∅ → (𝐴 ≠ ∅ → ((𝐴 × 𝐵) ∈ 𝑉 → 𝐵 ∈ V)))
12	5, 11	pm2.61ine 3023	1 ⊢ (𝐴 ≠ ∅ → ((𝐴 × 𝐵) ∈ 𝑉 → 𝐵 ∈ V))

Syntax hints:   → wi 4   ∧ wa 395   = wceq 1537   ∈ wcel 2106   ≠ wne 2938  Vcvv 3478  ∅c0 4339   × cxp 5687

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 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-11 2155  ax-12 2175  ax-ext 2706  ax-sep 5302  ax-nul 5312  ax-pr 5438  ax-un 7754

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1540  df-fal 1550  df-ex 1777  df-sb 2063  df-clab 2713  df-cleq 2727  df-clel 2814  df-ne 2939  df-ral 3060  df-rex 3069  df-rab 3434  df-v 3480  df-dif 3966  df-un 3968  df-in 3970  df-ss 3980  df-nul 4340  df-if 4532  df-sn 4632  df-pr 4634  df-op 4638  df-uni 4913  df-br 5149  df-opab 5211  df-xp 5695  df-rel 5696  df-cnv 5697  df-dm 5699  df-rn 5700

This theorem is referenced by:  bj-xpnzexb  36944