Theorem resopab2 6027

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 6025	. 2 ⊢ ({⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)} ↾ 𝐴) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝜑))}
2		ssel 3902	. . . . . 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 5170	. 2 ⊢ (𝐴 ⊆ 𝐵 → {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝜑))} = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)})
8	1, 7	eqtrid 2807	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 1569   ∈ wcel 2144   ⊆ wss 3895  {copab 5166   ↾ cres 5652

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-10 2177  ax-12 2212  ax-ext 2732  ax-sep 5248  ax-pr 5390

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-nf 1816  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-rab 3413  df-v 3452  df-ss 3896  df-dif 4047  df-in 4091  df-un 4151  df-nul 4279  df-if 4482  df-sn 4585  df-pr 4587  df-op 4835  df-opab 5167  df-xp 5656  df-rel 5657  df-res 5662

This theorem is used by:  resmpt  6028  marypha2lem4  9433