Theorem resopab2 6040

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 6038	. 2 ⊢ ({⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)} ↾ 𝐴) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝜑))}
2		ssel 3907	. . . . . 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 5174	. 2 ⊢ (𝐴 ⊆ 𝐵 → {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝜑))} = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)})
8	1, 7	eqtrid 2810	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 2147   ⊆ wss 3900  {copab 5170   ↾ cres 5657

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-10 2180  ax-12 2216  ax-ext 2735  ax-sep 5252  ax-pr 5395

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 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4839  df-opab 5171  df-xp 5661  df-rel 5662  df-res 5667

This theorem is used by:  resmpt  6041  marypha2lem4  9457