Theorem resopab2 5110

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 5107	. 2 ⊢ ({⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)} ↾ 𝐴) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝜑))}
2		ssel 3242	. . . . . 6 ⊢ (𝐴 ⊆ 𝐵 → (𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
3	2	pm4.71d 397	. . . . 5 ⊢ (𝐴 ⊆ 𝐵 → (𝑥 ∈ 𝐴 ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)))
4	3	anbi1d 469	. . . 4 ⊢ (𝐴 ⊆ 𝐵 → ((𝑥 ∈ 𝐴 ∧ 𝜑) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∧ 𝜑)))
5		anass 405	. . . 4 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∧ 𝜑) ↔ (𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝜑)))
6	4, 5	bitr2di 197	. . 3 ⊢ (𝐴 ⊆ 𝐵 → ((𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝜑)) ↔ (𝑥 ∈ 𝐴 ∧ 𝜑)))
7	6	opabbidv 4197	. 2 ⊢ (𝐴 ⊆ 𝐵 → {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ 𝜑))} = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)})
8	1, 7	eqtrid 2283	1 ⊢ (𝐴 ⊆ 𝐵 → ({⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)} ↾ 𝐴) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)})

This proof depends on syntax axioms:   → wi 4   ∧ wa 104   = wceq 1402   ∈ wcel 2209   ⊆ wss 3220  {copab 4191   ↾ cres 4776

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-14 2212  ax-ext 2220  ax-sep 4249  ax-pow 4311  ax-pr 4346

This proof depends on definitions:  df-bi 117  df-3an 1011  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-ral 2533  df-rex 2534  df-v 2823  df-un 3224  df-in 3226  df-ss 3233  df-pw 3690  df-sn 3715  df-pr 3716  df-op 3718  df-opab 4193  df-xp 4780  df-rel 4781  df-res 4786

This theorem is used by:  resmpt  5111