Theorem riinrab 4969

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 4967	. . 3 ⊢ (𝑋 = ∅ → (𝐴 ∩ ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑}) = 𝐴)
2		rzal 4411	. . . . 5 ⊢ (𝑋 = ∅ → ∀𝑥 ∈ 𝑋 𝜑)
3	2	ralrimivw 3150	. . . 4 ⊢ (𝑋 = ∅ → ∀𝑦 ∈ 𝐴 ∀𝑥 ∈ 𝑋 𝜑)
4		rabid2 3334	. . . 4 ⊢ (𝐴 = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑} ↔ ∀𝑦 ∈ 𝐴 ∀𝑥 ∈ 𝑋 𝜑)
5	3, 4	sylibr 237	. . 3 ⊢ (𝑋 = ∅ → 𝐴 = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑})
6	1, 5	eqtrd 2833	. 2 ⊢ (𝑋 = ∅ → (𝐴 ∩ ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑}) = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑})
7		ssrab2 4007	. . . . 5 ⊢ {𝑦 ∈ 𝐴 ∣ 𝜑} ⊆ 𝐴
8	7	rgenw 3118	. . . 4 ⊢ ∀𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑} ⊆ 𝐴
9		riinn0 4968	. . . 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 4954	. . 3 ⊢ (𝑋 ≠ ∅ → ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑} = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑})
12	10, 11	eqtrd 2833	. 2 ⊢ (𝑋 ≠ ∅ → (𝐴 ∩ ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑}) = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑})
13	6, 12	pm2.61ine 3070	1 ⊢ (𝐴 ∩ ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑}) = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑}

Syntax hints:   = wceq 1538   ≠ wne 2987  ∀wral 3106  {crab 3110   ∩ cin 3880   ⊆ wss 3881  ∅c0 4243  ∩ ciin 4882

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2113  ax-9 2121  ax-10 2142  ax-11 2158  ax-12 2175  ax-ext 2770

This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-tru 1541  df-ex 1782  df-nf 1786  df-sb 2070  df-clab 2777  df-cleq 2791  df-clel 2870  df-nfc 2938  df-ne 2988  df-ral 3111  df-rex 3112  df-rab 3115  df-v 3443  df-dif 3884  df-in 3888  df-ss 3898  df-nul 4244  df-iin 4884

This theorem is referenced by:  acsfn1  16924  acsfn1c  16925  acsfn2  16926  cntziinsn  18457  acsfn1p  19571  csscld  23853