Theorem inab 4258

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 2117	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))
2		df-clab 2741	. . 3 ⊢ (𝑦 ∈ {𝑥 ∣ (𝜑 ∧ 𝜓)} ↔ [𝑦 / 𝑥](𝜑 ∧ 𝜓))
3		df-clab 2741	. . . 4 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑)
4		df-clab 2741	. . . 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 4161	1 ⊢ ({𝑥 ∣ 𝜑} ∩ {𝑥 ∣ 𝜓}) = {𝑥 ∣ (𝜑 ∧ 𝜓)}

This proof depends on syntax axioms:   ∧ wa 401   = wceq 1570  [wsb 2099   ∈ wcel 2145  {cab 2740   ∩ cin 3901

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2741  df-cleq 2754  df-clel 2837  df-v 3455  df-in 3909

This theorem is used by:  inrab  4265  inrab2  4266  dfrab3  4268  orduniss2  7833  ssenen  9153  hashf1lem2  14525  symgsubmefmnd  19531  ballotlem2  35008  fmla0disjsuc  35985  dfiota3  36508  bj-inrab  37679  ptrest  38376  dmxrn  39143  ecqmap  39205  sticksstones22  43042  diophin  43625