Theorem inab 3499

Description: Intersection of two class abstractions. (Contributed by NM, 29-Sep-2002.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)

Assertion

Ref	Expression
inab	⊢ ({𝑥 ∣ 𝜑} ∩ {𝑥 ∣ 𝜓}) = {𝑥 ∣ (𝜑 ∧ 𝜓)}

Proof of Theorem inab

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		sban 2015	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))
2		df-clab 2225	. . 3 ⊢ (𝑦 ∈ {𝑥 ∣ (𝜑 ∧ 𝜓)} ↔ [𝑦 / 𝑥](𝜑 ∧ 𝜓))
3		df-clab 2225	. . . 4 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑)
4		df-clab 2225	. . . 4 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜓} ↔ [𝑦 / 𝑥]𝜓)
5	3, 4	anbi12i 464	. . 3 ⊢ ((𝑦 ∈ {𝑥 ∣ 𝜑} ∧ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))
6	1, 2, 5	3bitr4ri 213	. 2 ⊢ ((𝑦 ∈ {𝑥 ∣ 𝜑} ∧ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ 𝑦 ∈ {𝑥 ∣ (𝜑 ∧ 𝜓)})
7	6	ineqri 3424	1 ⊢ ({𝑥 ∣ 𝜑} ∩ {𝑥 ∣ 𝜓}) = {𝑥 ∣ (𝜑 ∧ 𝜓)}

Syntax hints:   ∧ wa 104   = wceq 1402  [wsb 1815   ∈ wcel 2209  {cab 2224   ∩ cin 3219

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 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This theorem depends on definitions:  df-bi 117  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-v 2823  df-in 3226

This theorem is referenced by:  inrab  3505  inrab2  3506  dfrab2  3508  dfrab3  3509  imainlem  5460  imain  5461  ssenen  7146  hashf1lem2  11269  ballotfilem2  13211