Theorem inab 4254

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 2116	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))
2		df-clab 2739	. . 3 ⊢ (𝑦 ∈ {𝑥 ∣ (𝜑 ∧ 𝜓)} ↔ [𝑦 / 𝑥](𝜑 ∧ 𝜓))
3		df-clab 2739	. . . 4 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑)
4		df-clab 2739	. . . 4 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜓} ↔ [𝑦 / 𝑥]𝜓)
5	3, 4	anbi12i 640	. . 3 ⊢ ((𝑦 ∈ {𝑥 ∣ 𝜑} ∧ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))
6	1, 2, 5	3bitr4ri 307	. 2 ⊢ ((𝑦 ∈ {𝑥 ∣ 𝜑} ∧ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ 𝑦 ∈ {𝑥 ∣ (𝜑 ∧ 𝜓)})
7	6	ineqri 4109	1 ⊢ ({𝑥 ∣ 𝜑} ∩ {𝑥 ∣ 𝜓}) = {𝑥 ∣ (𝜑 ∧ 𝜓)}

This proof depends on syntax axioms:   ∧ wa 401   = wceq 1569  [wsb 2098   ∈ wcel 2144  {cab 2738   ∩ cin 4089

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1572  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-v 3452  df-in 4091

This theorem is used by:  inrab  4261  inrab2  4262  dfrab3  4264  orduniss2  7844  ssenen  9173  hashf1lem2  14604  symgsubmefmnd  19617  ballotlem2  35167  fmla0disjsuc  36249  dfiota3  36777  bj-inrab  37932  ptrest  38630  dmxrn  39412  ecqmap  39474  sticksstones22  43311  diophin  43876