Theorem rabdif 4269

Description: Move difference in and out of a restricted class abstraction. (Contributed by Steven Nguyen, 6-Jun-2023.)

Assertion

Ref	Expression
rabdif	⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∖ 𝐵) = {𝑥 ∈ (𝐴 ∖ 𝐵) ∣ 𝜑}

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵

Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem rabdif

Step	Hyp	Ref	Expression
1		indif2 4229	. 2 ⊢ ({𝑥 ∣ 𝜑} ∩ (𝐴 ∖ 𝐵)) = (({𝑥 ∣ 𝜑} ∩ 𝐴) ∖ 𝐵)
2		dfrab2 4268	. 2 ⊢ {𝑥 ∈ (𝐴 ∖ 𝐵) ∣ 𝜑} = ({𝑥 ∣ 𝜑} ∩ (𝐴 ∖ 𝐵))
3		dfrab2 4268	. . 3 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = ({𝑥 ∣ 𝜑} ∩ 𝐴)
4	3	difeq1i 4062	. 2 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∖ 𝐵) = (({𝑥 ∣ 𝜑} ∩ 𝐴) ∖ 𝐵)
5	1, 2, 4	3eqtr4ri 2797	1 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∖ 𝐵) = {𝑥 ∈ (𝐴 ∖ 𝐵) ∣ 𝜑}

This proof depends on syntax axioms:   = wceq 1570  {cab 2741  {crab 3416   ∖ cdif 4048   ∩ 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-rab 3417  df-v 3456  df-dif 4050  df-in 4094

This theorem is used by:  prjspeclsp  43906  dfclnbgr4  49195  clnbupgr  49204  dfnbgr5  49222  dfnbgr6  49228