Theorem resopab2 6030

Description: Restriction of a class abstraction of ordered pairs. (Contributed by NM, 24-Aug-2007.)

Assertion

Ref	Expression
resopab2	⊢ (𝐴 ⊆ 𝐵 → ({⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)} ↾ 𝐴) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)})

Distinct variable groups:   𝑥,𝑦,𝐴   𝑥,𝐵,𝑦

Allowed substitution hints:   𝜑(𝑥, 𝑦)

Proof of Theorem resopab2

Step	Hyp	Ref	Expression
1		resopab 6028	. 2 ⊢ ({⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)} ↾ 𝐴) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝜑))}
2		ssel 3925	. . . . . 6 ⊢ (𝐴 ⊆ 𝐵 → (𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
3	2	pm4.71d 571	. . . . 5 ⊢ (𝐴 ⊆ 𝐵 → (𝑥 ∈ 𝐴 ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)))
4	3	anbi1d 643	. . . 4 ⊢ (𝐴 ⊆ 𝐵 → ((𝑥 ∈ 𝐴 ∧ 𝜑) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∧ 𝜑)))
5		anass 474	. . . 4 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∧ 𝜑) ↔ (𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝜑)))
6	4, 5	bitr2di 291	. . 3 ⊢ (𝐴 ⊆ 𝐵 → ((𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝜑)) ↔ (𝑥 ∈ 𝐴 ∧ 𝜑)))
7	6	opabbidv 5171	. 2 ⊢ (𝐴 ⊆ 𝐵 → {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝜑))} = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)})
8	1, 7	eqtrid 2808	1 ⊢ (𝐴 ⊆ 𝐵 → ({⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)} ↾ 𝐴) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)})

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   = wceq 1570   ∈ wcel 2145   ⊆ wss 3899  {copab 5167   ↾ cres 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-10 2178  ax-12 2213  ax-ext 2733  ax-sep 5249  ax-pr 5391

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-nf 1817  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-rab 3414  df-v 3453  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 5657  df-rel 5658  df-res 5663

This theorem is used by:  resmpt  6031  marypha2lem4  9414