Theorem riinrab 5009

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 5007	. . 3 ⊢ (𝑋 = ∅ → (𝐴 ∩ ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑}) = 𝐴)
2		rzal 4436	. . . . 5 ⊢ (𝑋 = ∅ → ∀𝑥 ∈ 𝑋 𝜑)
3	2	ralrimivw 3108	. . . 4 ⊢ (𝑋 = ∅ → ∀𝑦 ∈ 𝐴 ∀𝑥 ∈ 𝑋 𝜑)
4		rabid2 3307	. . . 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 2778	. 2 ⊢ (𝑋 = ∅ → (𝐴 ∩ ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑}) = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑})
7		ssrab2 4009	. . . . 5 ⊢ {𝑦 ∈ 𝐴 ∣ 𝜑} ⊆ 𝐴
8	7	rgenw 3075	. . . 4 ⊢ ∀𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑} ⊆ 𝐴
9		riinn0 5008	. . . 4 ⊢ ((∀𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑} ⊆ 𝐴 ∧ 𝑋 ≠ ∅) → (𝐴 ∩ ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑}) = ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑})
10	8, 9	mpan 686	. . 3 ⊢ (𝑋 ≠ ∅ → (𝐴 ∩ ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑}) = ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑})
11		iinrab 4994	. . 3 ⊢ (𝑋 ≠ ∅ → ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑} = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑})
12	10, 11	eqtrd 2778	. 2 ⊢ (𝑋 ≠ ∅ → (𝐴 ∩ ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑}) = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑})
13	6, 12	pm2.61ine 3027	1 ⊢ (𝐴 ∩ ∩ 𝑥 ∈ 𝑋 {𝑦 ∈ 𝐴 ∣ 𝜑}) = {𝑦 ∈ 𝐴 ∣ ∀𝑥 ∈ 𝑋 𝜑}

Syntax hints:   = wceq 1539   ≠ wne 2942  ∀wral 3063  {crab 3067   ∩ cin 3882   ⊆ wss 3883  ∅c0 4253  ∩ ciin 4922

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-10 2139  ax-11 2156  ax-12 2173  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-tru 1542  df-fal 1552  df-ex 1784  df-nf 1788  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-nfc 2888  df-ne 2943  df-ral 3068  df-rex 3069  df-rab 3072  df-v 3424  df-dif 3886  df-in 3890  df-ss 3900  df-nul 4254  df-iin 4924

This theorem is referenced by:  acsfn1  17287  acsfn1c  17288  acsfn2  17289  cntziinsn  18856  acsfn1p  19982  csscld  24318