Theorem inrab 3435

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 2484	. . 3 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)}
2		df-rab 2484	. . 3 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)}
3	1, 2	ineq12i 3362	. 2 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐴 ∣ 𝜓}) = ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∩ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)})
4		df-rab 2484	. . 3 ⊢ {𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ (𝜑 ∧ 𝜓))}
5		inab 3431	. . . 4 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∩ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)}) = {𝑥 ∣ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (𝑥 ∈ 𝐴 ∧ 𝜓))}
6		anandi 590	. . . . 5 ⊢ ((𝑥 ∈ 𝐴 ∧ (𝜑 ∧ 𝜓)) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (𝑥 ∈ 𝐴 ∧ 𝜓)))
7	6	abbii 2312	. . . 4 ⊢ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ (𝜑 ∧ 𝜓))} = {𝑥 ∣ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (𝑥 ∈ 𝐴 ∧ 𝜓))}
8	5, 7	eqtr4i 2220	. . 3 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∩ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)}) = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ (𝜑 ∧ 𝜓))}
9	4, 8	eqtr4i 2220	. 2 ⊢ {𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)} = ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∩ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)})
10	3, 9	eqtr4i 2220	1 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐴 ∣ 𝜓}) = {𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)}

Syntax hints:   ∧ wa 104   = wceq 1364   ∈ wcel 2167  {cab 2182  {crab 2479   ∩ cin 3156

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 710  ax-5 1461  ax-7 1462  ax-gen 1463  ax-ie1 1507  ax-ie2 1508  ax-8 1518  ax-10 1519  ax-11 1520  ax-i12 1521  ax-bndl 1523  ax-4 1524  ax-17 1540  ax-i9 1544  ax-ial 1548  ax-i5r 1549  ax-ext 2178

This theorem depends on definitions:  df-bi 117  df-tru 1367  df-nf 1475  df-sb 1777  df-clab 2183  df-cleq 2189  df-clel 2192  df-nfc 2328  df-rab 2484  df-v 2765  df-in 3163

This theorem is referenced by:  rabnc  3483  iooinsup  11442  phiprmpw  12390  unennn  12614  dfrhm2  13710  lgsquadlem2  15319