Theorem inab 4257

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 2119	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))
2		df-clab 2742	. . 3 ⊢ (𝑦 ∈ {𝑥 ∣ (𝜑 ∧ 𝜓)} ↔ [𝑦 / 𝑥](𝜑 ∧ 𝜓))
3		df-clab 2742	. . . 4 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑)
4		df-clab 2742	. . . 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 4112	1 ⊢ ({𝑥 ∣ 𝜑} ∩ {𝑥 ∣ 𝜓}) = {𝑥 ∣ (𝜑 ∧ 𝜓)}

This proof depends on syntax axioms:   ∧ wa 401   = wceq 1570  [wsb 2101   ∈ wcel 2147  {cab 2741   ∩ cin 4092

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-v 3456  df-in 4094

This theorem is used by:  inrab  4264  inrab2  4265  dfrab3  4267  orduniss2  7871  ssenen  9201  hashf1lem2  14631  symgsubmefmnd  19645  ballotlem2  35204  fmla0disjsuc  36343  dfiota3  36876  bj-inrab  38031  ptrest  38742  dmxrn  39522  ecqmap  39584  sticksstones22  43421  diophin  43987