Theorem bj-xpnzex 37319

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 7867 (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 5236	. . . . 5 ⊢ ∅ ∈ V
2		eleq1a 2835	. . . . 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 6117	. . . 4 ⊢ ((𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅) ↔ (𝐴 × 𝐵) ≠ ∅)
7		xpexr2 7866	. . . . . 6 ⊢ (((𝐴 × 𝐵) ∈ 𝑉 ∧ (𝐴 × 𝐵) ≠ ∅) → (𝐴 ∈ V ∧ 𝐵 ∈ V))
8	7	simprd 496	. . . . 5 ⊢ (((𝐴 × 𝐵) ∈ 𝑉 ∧ (𝐴 × 𝐵) ≠ ∅) → 𝐵 ∈ V)
9	8	expcom 414	. . . 4 ⊢ ((𝐴 × 𝐵) ≠ ∅ → ((𝐴 × 𝐵) ∈ 𝑉 → 𝐵 ∈ V))
10	6, 9	sylbi 218	. . 3 ⊢ ((𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅) → ((𝐴 × 𝐵) ∈ 𝑉 → 𝐵 ∈ V))
11	10	expcom 414	. 2 ⊢ (𝐵 ≠ ∅ → (𝐴 ≠ ∅ → ((𝐴 × 𝐵) ∈ 𝑉 → 𝐵 ∈ V)))
12	5, 11	pm2.61ine 3018	1 ⊢ (𝐴 ≠ ∅ → ((𝐴 × 𝐵) ∈ 𝑉 → 𝐵 ∈ V))

Syntax hints:   → wi 4   ∧ wa 396   = wceq 1547   ∈ wcel 2119   ≠ wne 2935  Vcvv 3432  ∅c0 4268   × cxp 5623

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-11 2168  ax-ext 2712  ax-sep 5225  ax-nul 5235  ax-pr 5369  ax-un 7685

This theorem depends on definitions:  df-bi 208  df-an 397  df-or 854  df-3an 1094  df-tru 1550  df-fal 1560  df-ex 1787  df-sb 2074  df-clab 2719  df-cleq 2732  df-clel 2815  df-ne 2936  df-ral 3055  df-rex 3065  df-rab 3393  df-v 3434  df-dif 3893  df-un 3895  df-in 3897  df-ss 3907  df-nul 4269  df-if 4462  df-sn 4563  df-pr 4565  df-op 4569  df-uni 4846  df-br 5080  df-opab 5142  df-xp 5631  df-rel 5632  df-cnv 5633  df-dm 5635  df-rn 5636

This theorem is referenced by:  bj-xpnzexb  37321