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Theorem bj-xpnzex 32620
 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 7062 (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 4755 . . . . 5 ∅ ∈ V
2 eleq1a 2693 . . . . 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 5517 . . . 4 ((𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅) ↔ (𝐴 × 𝐵) ≠ ∅)
7 xpexr2 7061 . . . . . 6 (((𝐴 × 𝐵) ∈ 𝑉 ∧ (𝐴 × 𝐵) ≠ ∅) → (𝐴 ∈ V ∧ 𝐵 ∈ V))
87simprd 479 . . . . 5 (((𝐴 × 𝐵) ∈ 𝑉 ∧ (𝐴 × 𝐵) ≠ ∅) → 𝐵 ∈ V)
98expcom 451 . . . 4 ((𝐴 × 𝐵) ≠ ∅ → ((𝐴 × 𝐵) ∈ 𝑉𝐵 ∈ V))
106, 9sylbi 207 . . 3 ((𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅) → ((𝐴 × 𝐵) ∈ 𝑉𝐵 ∈ V))
1110expcom 451 . 2 (𝐵 ≠ ∅ → (𝐴 ≠ ∅ → ((𝐴 × 𝐵) ∈ 𝑉𝐵 ∈ V)))
125, 11pm2.61ine 2873 1 (𝐴 ≠ ∅ → ((𝐴 × 𝐵) ∈ 𝑉𝐵 ∈ V))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ∧ wa 384   = wceq 1480   ∈ wcel 1987   ≠ wne 2790  Vcvv 3189  ∅c0 3896   × cxp 5077 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1719  ax-4 1734  ax-5 1836  ax-6 1885  ax-7 1932  ax-8 1989  ax-9 1996  ax-10 2016  ax-11 2031  ax-12 2044  ax-13 2245  ax-ext 2601  ax-sep 4746  ax-nul 4754  ax-pr 4872  ax-un 6909 This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-3an 1038  df-tru 1483  df-ex 1702  df-nf 1707  df-sb 1878  df-eu 2473  df-mo 2474  df-clab 2608  df-cleq 2614  df-clel 2617  df-nfc 2750  df-ne 2791  df-ral 2912  df-rex 2913  df-rab 2916  df-v 3191  df-dif 3562  df-un 3564  df-in 3566  df-ss 3573  df-nul 3897  df-if 4064  df-sn 4154  df-pr 4156  df-op 4160  df-uni 4408  df-br 4619  df-opab 4679  df-xp 5085  df-rel 5086  df-cnv 5087  df-dm 5089  df-rn 5090 This theorem is referenced by:  bj-xpnzexb  32622
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