Theorem relopabiALT 5804

Description: Alternate proof of relopabi 5803 (shorter but uses more axioms). (Contributed by Mario Carneiro, 21-Dec-2013.) (Proof modification is discouraged.) (New usage is discouraged.)

Hypothesis

Ref	Expression
relopabi.1	⊢ 𝐴 = {⟨𝑥, 𝑦⟩ ∣ 𝜑}

Assertion

Ref	Expression
relopabiALT	⊢ Rel 𝐴

Proof of Theorem relopabiALT

Dummy variable 𝑧 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		relopabi.1	. . . 4 ⊢ 𝐴 = {⟨𝑥, 𝑦⟩ ∣ 𝜑}
2		df-opab 5168	. . . 4 ⊢ {⟨𝑥, 𝑦⟩ ∣ 𝜑} = {𝑧 ∣ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑)}
3	1, 2	eqtri 2783	. . 3 ⊢ 𝐴 = {𝑧 ∣ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑)}
4		vex 3454	. . . . . . . 8 ⊢ 𝑥 ∈ V
5		vex 3454	. . . . . . . 8 ⊢ 𝑦 ∈ V
6	4, 5	opelvv 5695	. . . . . . 7 ⊢ ⟨𝑥, 𝑦⟩ ∈ (V × V)
7		eleq1 2848	. . . . . . 7 ⊢ (𝑧 = ⟨𝑥, 𝑦⟩ → (𝑧 ∈ (V × V) ↔ ⟨𝑥, 𝑦⟩ ∈ (V × V)))
8	6, 7	mpbiri 261	. . . . . 6 ⊢ (𝑧 = ⟨𝑥, 𝑦⟩ → 𝑧 ∈ (V × V))
9	8	adantr 486	. . . . 5 ⊢ ((𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) → 𝑧 ∈ (V × V))
10	9	exlimivv 1965	. . . 4 ⊢ (∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) → 𝑧 ∈ (V × V))
11	10	abssi 4016	. . 3 ⊢ {𝑧 ∣ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑)} ⊆ (V × V)
12	3, 11	eqsstri 3977	. 2 ⊢ 𝐴 ⊆ (V × V)
13		df-rel 5662	. 2 ⊢ (Rel 𝐴 ↔ 𝐴 ⊆ (V × V))
14	12, 13	mpbir 234	1 ⊢ Rel 𝐴

This proof depends on syntax axioms:   ∧ wa 401   = wceq 1570  ∃wex 1812   ∈ wcel 2145  {cab 2738  Vcvv 3450   ⊆ wss 3899  ⟨cop 4590  {copab 5167   × cxp 5653  Rel wrel 5660

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 2732  ax-sep 5251  ax-pr 5398

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 2739  df-cleq 2752  df-clel 2835  df-ral 3077  df-rex 3087  df-rab 3413  df-v 3452  df-dif 3902  df-un 3904  df-in 3906  df-ss 3916  df-nul 4280  df-if 4483  df-sn 4585  df-pr 4587  df-op 4591  df-opab 5168  df-xp 5661  df-rel 5662

This theorem is used by: (None)