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Theorem bj-xpnzex 32139
 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 exported form (curried form) of xpexcnv 7001 (up to commutation in the product). (Contributed by BJ, 6-Oct-2018.) (Proof modification is discouraged.)
Assertion
Ref Expression
bj-xpnzex (𝐴 ≠ ∅ → ((𝐴 × 𝐵) ∈ 𝑉𝐵 ∈ V))

Proof of Theorem bj-xpnzex
StepHypRef Expression
1 0ex 4718 . . . . 5 ∅ ∈ V
2 eleq1a 2683 . . . . 5 (∅ ∈ V → (𝐵 = ∅ → 𝐵 ∈ V))
31, 2ax-mp 5 . . . 4 (𝐵 = ∅ → 𝐵 ∈ V)
43a1d 25 . . 3 (𝐵 = ∅ → ((𝐴 × 𝐵) ∈ 𝑉𝐵 ∈ V))
54a1d 25 . 2 (𝐵 = ∅ → (𝐴 ≠ ∅ → ((𝐴 × 𝐵) ∈ 𝑉𝐵 ∈ V)))
6 xpnz 5472 . . . 4 ((𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅) ↔ (𝐴 × 𝐵) ≠ ∅)
7 xpexr2 7000 . . . . . 6 (((𝐴 × 𝐵) ∈ 𝑉 ∧ (𝐴 × 𝐵) ≠ ∅) → (𝐴 ∈ V ∧ 𝐵 ∈ V))
87simprd 478 . . . . 5 (((𝐴 × 𝐵) ∈ 𝑉 ∧ (𝐴 × 𝐵) ≠ ∅) → 𝐵 ∈ V)
98expcom 450 . . . 4 ((𝐴 × 𝐵) ≠ ∅ → ((𝐴 × 𝐵) ∈ 𝑉𝐵 ∈ V))
106, 9sylbi 206 . . 3 ((𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅) → ((𝐴 × 𝐵) ∈ 𝑉𝐵 ∈ V))
1110expcom 450 . 2 (𝐵 ≠ ∅ → (𝐴 ≠ ∅ → ((𝐴 × 𝐵) ∈ 𝑉𝐵 ∈ V)))
125, 11pm2.61ine 2865 1 (𝐴 ≠ ∅ → ((𝐴 × 𝐵) ∈ 𝑉𝐵 ∈ V))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ∧ wa 383   = wceq 1475   ∈ wcel 1977   ≠ wne 2780  Vcvv 3173  ∅c0 3874   × cxp 5036 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-8 1979  ax-9 1986  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590  ax-sep 4709  ax-nul 4717  ax-pr 4833  ax-un 6847 This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-3an 1033  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-eu 2462  df-mo 2463  df-clab 2597  df-cleq 2603  df-clel 2606  df-nfc 2740  df-ne 2782  df-ral 2901  df-rex 2902  df-rab 2905  df-v 3175  df-dif 3543  df-un 3545  df-in 3547  df-ss 3554  df-nul 3875  df-if 4037  df-sn 4126  df-pr 4128  df-op 4132  df-uni 4373  df-br 4584  df-opab 4644  df-xp 5044  df-rel 5045  df-cnv 5046  df-dm 5048  df-rn 5049 This theorem is referenced by:  bj-xpnzexb  32141
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