Theorem bj-xpnzex 37711

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 7921 (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 5268	. . . . 5 ⊢ ∅ ∈ V
2		eleq1a 2857	. . . . 5 ⊢ (∅ ∈ V → (𝐵 = ∅ → 𝐵 ∈ V))
3	1, 2	ax-mp 5	. . . 4 ⊢ (𝐵 = ∅ → 𝐵 ∈ V)
4	3	a1d 26	. . 3 ⊢ (𝐵 = ∅ → ((𝐴 × 𝐵) ∈ 𝑉 → 𝐵 ∈ V))
5	4	a1d 26	. 2 ⊢ (𝐵 = ∅ → (𝐴 ≠ ∅ → ((𝐴 × 𝐵) ∈ 𝑉 → 𝐵 ∈ V)))
6		xpnz 6155	. . . 4 ⊢ ((𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅) ↔ (𝐴 × 𝐵) ≠ ∅)
7		xpexr2 7920	. . . . . 6 ⊢ (((𝐴 × 𝐵) ∈ 𝑉 ∧ (𝐴 × 𝐵) ≠ ∅) → (𝐴 ∈ V ∧ 𝐵 ∈ V))
8	7	simprd 501	. . . . 5 ⊢ (((𝐴 × 𝐵) ∈ 𝑉 ∧ (𝐴 × 𝐵) ≠ ∅) → 𝐵 ∈ V)
9	8	expcom 419	. . . 4 ⊢ ((𝐴 × 𝐵) ≠ ∅ → ((𝐴 × 𝐵) ∈ 𝑉 → 𝐵 ∈ V))
10	6, 9	sylbi 220	. . 3 ⊢ ((𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅) → ((𝐴 × 𝐵) ∈ 𝑉 → 𝐵 ∈ V))
11	10	expcom 419	. 2 ⊢ (𝐵 ≠ ∅ → (𝐴 ≠ ∅ → ((𝐴 × 𝐵) ∈ 𝑉 → 𝐵 ∈ V)))
12	5, 11	pm2.61ine 3040	1 ⊢ (𝐴 ≠ ∅ → ((𝐴 × 𝐵) ∈ 𝑉 → 𝐵 ∈ V))

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   = wceq 1570   ∈ wcel 2145   ≠ wne 2957  Vcvv 3453  ∅c0 4282   × cxp 5657

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2734  ax-sep 5255  ax-nul 5267  ax-pr 5402  ax-un 7740

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2741  df-cleq 2754  df-clel 2837  df-ne 2958  df-ral 3079  df-rex 3089  df-rab 3415  df-v 3455  df-dif 3905  df-un 3907  df-in 3909  df-ss 3919  df-nul 4283  df-if 4486  df-sn 4588  df-pr 4590  df-op 4594  df-uni 4871  df-br 5108  df-opab 5172  df-xp 5665  df-rel 5666  df-cnv 5667  df-dm 5669  df-rn 5670

This theorem is used by:  bj-xpnzexb  37713