Theorem bj-opabssvv 37902

Description: A variant of relopabiv 5801 (which could be proved from it, similarly to relxp 5673 from xpss 5671). (Contributed by BJ, 28-Dec-2023.)

Assertion

Ref	Expression
bj-opabssvv	⊢ {⟨𝑥, 𝑦⟩ ∣ 𝜑} ⊆ (V × V)

Distinct variable group:   𝑥,𝑦

Allowed substitution hints:   𝜑(𝑥, 𝑦)

Proof of Theorem bj-opabssvv

Step	Hyp	Ref	Expression
1		vex 3454	. . . . 5 ⊢ 𝑥 ∈ V
2		vex 3454	. . . . 5 ⊢ 𝑦 ∈ V
3	1, 2	pm3.2i 476	. . . 4 ⊢ (𝑥 ∈ V ∧ 𝑦 ∈ V)
4	3	a1i 11	. . 3 ⊢ (𝜑 → (𝑥 ∈ V ∧ 𝑦 ∈ V))
5	4	ssopab2i 5529	. 2 ⊢ {⟨𝑥, 𝑦⟩ ∣ 𝜑} ⊆ {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ V ∧ 𝑦 ∈ V)}
6		df-xp 5661	. 2 ⊢ (V × V) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ V ∧ 𝑦 ∈ V)}
7	5, 6	sseqtrri 3980	1 ⊢ {⟨𝑥, 𝑦⟩ ∣ 𝜑} ⊆ (V × V)

This proof depends on syntax axioms:   ∧ wa 401   ∈ wcel 2145  Vcvv 3450   ⊆ wss 3899  {copab 5167   × cxp 5653

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

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-v 3452  df-ss 3916  df-opab 5168  df-xp 5661

This theorem is used by: (None)