Theorem inrab 3528

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 2624	. . 3 ⊢ {x ∈ A ∣ φ} = {x ∣ (x ∈ A ∧ φ)}
2		df-rab 2624	. . 3 ⊢ {x ∈ A ∣ ψ} = {x ∣ (x ∈ A ∧ ψ)}
3	1, 2	ineq12i 3456	. 2 ⊢ ({x ∈ A ∣ φ} ∩ {x ∈ A ∣ ψ}) = ({x ∣ (x ∈ A ∧ φ)} ∩ {x ∣ (x ∈ A ∧ ψ)})
4		df-rab 2624	. . 3 ⊢ {x ∈ A ∣ (φ ∧ ψ)} = {x ∣ (x ∈ A ∧ (φ ∧ ψ))}
5		inab 3523	. . . 4 ⊢ ({x ∣ (x ∈ A ∧ φ)} ∩ {x ∣ (x ∈ A ∧ ψ)}) = {x ∣ ((x ∈ A ∧ φ) ∧ (x ∈ A ∧ ψ))}
6		anandi 801	. . . . 5 ⊢ ((x ∈ A ∧ (φ ∧ ψ)) ↔ ((x ∈ A ∧ φ) ∧ (x ∈ A ∧ ψ)))
7	6	abbii 2466	. . . 4 ⊢ {x ∣ (x ∈ A ∧ (φ ∧ ψ))} = {x ∣ ((x ∈ A ∧ φ) ∧ (x ∈ A ∧ ψ))}
8	5, 7	eqtr4i 2376	. . 3 ⊢ ({x ∣ (x ∈ A ∧ φ)} ∩ {x ∣ (x ∈ A ∧ ψ)}) = {x ∣ (x ∈ A ∧ (φ ∧ ψ))}
9	4, 8	eqtr4i 2376	. 2 ⊢ {x ∈ A ∣ (φ ∧ ψ)} = ({x ∣ (x ∈ A ∧ φ)} ∩ {x ∣ (x ∈ A ∧ ψ)})
10	3, 9	eqtr4i 2376	1 ⊢ ({x ∈ A ∣ φ} ∩ {x ∈ A ∣ ψ}) = {x ∈ A ∣ (φ ∧ ψ)}

This proof depends on syntax axioms:   ∧ wa 358   = wceq 1642   ∈ wcel 1710  {cab 2339  {crab 2619   ∩ cin 3209

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This proof depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-rab 2624  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214

This theorem is used by:  rabnc  3575