Theorem inrab 3236

Description: Intersection of two restricted class abstractions. (Contributed by NM, 1-Sep-2006.)

Assertion

Ref	Expression
inrab	⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐴 ∣ 𝜓}) = {𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)}

Proof of Theorem inrab

Step	Hyp	Ref	Expression
1		df-rab 2357	. . 3 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)}
2		df-rab 2357	. . 3 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)}
3	1, 2	ineq12i 3165	. 2 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐴 ∣ 𝜓}) = ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∩ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)})
4		df-rab 2357	. . 3 ⊢ {𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ (𝜑 ∧ 𝜓))}
5		inab 3232	. . . 4 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∩ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)}) = {𝑥 ∣ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (𝑥 ∈ 𝐴 ∧ 𝜓))}
6		anandi 554	. . . . 5 ⊢ ((𝑥 ∈ 𝐴 ∧ (𝜑 ∧ 𝜓)) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (𝑥 ∈ 𝐴 ∧ 𝜓)))
7	6	abbii 2194	. . . 4 ⊢ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ (𝜑 ∧ 𝜓))} = {𝑥 ∣ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (𝑥 ∈ 𝐴 ∧ 𝜓))}
8	5, 7	eqtr4i 2104	. . 3 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∩ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)}) = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ (𝜑 ∧ 𝜓))}
9	4, 8	eqtr4i 2104	. 2 ⊢ {𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)} = ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∩ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)})
10	3, 9	eqtr4i 2104	1 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐴 ∣ 𝜓}) = {𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)}

Syntax hints:   ∧ wa 102   = wceq 1284   ∈ wcel 1433  {cab 2067  {crab 2352   ∩ cin 2972

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-bndl 1439  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063

This theorem depends on definitions:  df-bi 115  df-tru 1287  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-nfc 2208  df-rab 2357  df-v 2603  df-in 2979

This theorem is referenced by:  rabnc  3277