Theorem riinrab 5081

Description: Relative intersection of a relative abstraction. (Contributed by Stefan O'Rear, 3-Apr-2015.)

Assertion

Ref	Expression
riinrab	⊢ (𝐴 ∩ ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑}) = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑}

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

Allowed substitution hints:   𝜑(𝑥,𝑦)

Proof of Theorem riinrab

Step	Hyp	Ref	Expression
1		riin0 5079	. . 3 ⊢ (𝑋 = ∅ → (𝐴 ∩ ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑}) = 𝐴)
2		rzal 4504	. . . . 5 ⊢ (𝑋 = ∅ → ∀𝑥 ∈ 𝑋 𝜑)
3	2	ralrimivw 3145	. . . 4 ⊢ (𝑋 = ∅ → ∀𝑦 ∈ 𝐴 ∀𝑥 ∈ 𝑋 𝜑)
4		rabid2 3459	. . . 4 ⊢ (𝐴 = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑} ↔ ∀𝑦 ∈ 𝐴 ∀𝑥 ∈ 𝑋 𝜑)
5	3, 4	sylibr 233	. . 3 ⊢ (𝑋 = ∅ → 𝐴 = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑})
6	1, 5	eqtrd 2767	. 2 ⊢ (𝑋 = ∅ → (𝐴 ∩ ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑}) = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑})
7		ssrab2 4073	. . . . 5 ⊢ {𝑦 ∈ 𝐴 ∣ 𝜑} ⊆ 𝐴
8	7	rgenw 3060	. . . 4 ⊢ ∀𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑} ⊆ 𝐴
9		riinn0 5080	. . . 4 ⊢ ((∀𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑} ⊆ 𝐴 ∧ 𝑋 ≠ ∅) → (𝐴 ∩ ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑}) = ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑})
10	8, 9	mpan 689	. . 3 ⊢ (𝑋 ≠ ∅ → (𝐴 ∩ ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑}) = ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑})
11		iinrab 5066	. . 3 ⊢ (𝑋 ≠ ∅ → ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑} = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑})
12	10, 11	eqtrd 2767	. 2 ⊢ (𝑋 ≠ ∅ → (𝐴 ∩ ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑}) = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑})
13	6, 12	pm2.61ine 3020	1 ⊢ (𝐴 ∩ ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑}) = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑}

Syntax hints:   = wceq 1534   ≠ wne 2935  ∀wral 3056  {crab 3427   ∩ cin 3943   ⊆ wss 3944  ∅c0 4318  ∩ ciin 4992

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004  ax-8 2101  ax-9 2109  ax-10 2130  ax-11 2147  ax-12 2164  ax-ext 2698

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 847  df-tru 1537  df-fal 1547  df-ex 1775  df-nf 1779  df-sb 2061  df-clab 2705  df-cleq 2719  df-clel 2805  df-nfc 2880  df-ne 2936  df-ral 3057  df-rex 3066  df-rab 3428  df-v 3471  df-dif 3947  df-in 3951  df-ss 3961  df-nul 4319  df-iin 4994

This theorem is referenced by:  acsfn1  17632  acsfn1c  17633  acsfn2  17634  cntziinsn  19279  acsfn1p  20676  csscld  25164