Theorem relopabiALT 5802

Description: Alternate proof of relopabi 5801 (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 5167	. . . 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 5691	. . . . . . 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 1964	. . . 4 ⊢ (∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) → 𝑧 ∈ (V × V))
11	10	abssi 3992	. . 3 ⊢ {𝑧 ∣ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑)} ⊆ (V × V)
12	3, 11	eqsstri 3955	. 2 ⊢ 𝐴 ⊆ (V × V)
13		df-rel 5658	. 2 ⊢ (Rel 𝐴 ↔ 𝐴 ⊆ (V × V))
14	12, 13	mpbir 234	1 ⊢ Rel 𝐴

This proof depends on syntax axioms:   ∧ wa 401   = wceq 1569  ∃wex 1811   ∈ wcel 2144  {cab 2738  Vcvv 3450   ⊆ wss 3896  ⟨cop 4834  {copab 5166   × cxp 5649  Rel wrel 5656

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732  ax-sep 5248  ax-pr 5391

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-ral 3077  df-rex 3087  df-rab 3413  df-v 3452  df-ss 3897  df-dif 4046  df-in 4090  df-un 4150  df-nul 4278  df-if 4482  df-sn 4585  df-pr 4587  df-op 4835  df-opab 5167  df-xp 5657  df-rel 5658

This theorem is used by: (None)